It is a bit annoying that, while it is well-known, there does not seem to be a complete proof of this result anywhere, while most pieces of the proof are in the literature. Mike Miller's answer is also missing details.
In what follows, an $n$-dimensional manifold is a 2nd countable Hausdorff space
locally homeomorphic to $R^n$. (In particular, by default, manifolds are not assumed to have boundary.)
I will start with a general definition:
Definition. A topological $n$-manifold $M$ is said to be tame if there exists a compact $n$-manifold with boundary $\bar{M}$ such that $M$ is homeomorphic to the interior of $\bar{M}$, i.e. $\bar{M} -\partial \bar{M}$.
In this case, the inclusion $M\to \bar{M}$ is a homotopy-equivalence. In particular (if $M$ is connected), $\pi_1(M)\cong \pi_1(\bar{M})$ is finitely-presented. Thus, every connected tame manifold has finitely presented fundamental group. The converse is famously false already in dimension 3: The Whitehead manifold is contractible but is not tame. In what follows, will be using homology with $Z_2$-coefficients.
The situation in the case of surfaces, on the other hand, is much simpler:
Theorem 1. Suppose that $M$ is a noncompact connected surface. Then the following are equivalent:
$M$ is tame.
$H_1(M):=H_1(M; Z_2)$ is finitely generated (i.e. is finite).
$\pi_1(M)$ is finitely-generated.
$\pi_1(M)$ is finitely-presented.
$\pi_1(M)$ is free of finite rank.
Proof. The implications (5)$\Rightarrow$(4)$\Rightarrow$(3)$\Rightarrow$(2) are immediate (the last one uses the Hurewicz theorem). The implication (1)$\Rightarrow$(5) follows from: If $M$ is tame then $\pi_1(M)\cong \pi_1(\bar{M})$. Since $M$ is assumed to be noncompact, $\bar{M}$ has nonempty boundary, hence, by applying inductively the van Kampen theorem, we conclude that $\pi_1(\bar{M})$ is free. Alternatively, one can use the much harder result that fundamental groups of connected noncompact surfaces are free.
Thus, it remains to prove the implication (2)$\Rightarrow$(1).
Since all surfaces admit triangulations, we can assume that $M$ is triangulated. Let $M^1$ denote the 1-skeleton of the triangulation.
Then $H_1(M^1)$ maps onto $H_1(M)$. Since $H_1(M)$ is assumed to be finitely generated and $M$ is connected, there is a finite connected subgraph $X\subset M^1$ such that the inclusion maps $X\to M^1\to M$ induce a surjective map $H_1(X)\to H_1(M)$. Let $Y\subset M$ be the regular neighborhood of $X$ in $M$. It is a compact surface with boundary. The inclusion $X\to Y$ is also a homotopy-equivalence, hence, the natural homomorphism $H_1(Y)\to H_1(M)$ is also surjective. Consider a boundary component $C$ of $Y$. If $C$ does not separate $M$ then, applying the Mayer-Vietoris sequence (or duality), we conclude that $H_1(X)\to H_1(M)$ is not surjective, which is impossible. Thus, $C$ separates $M$ in two components, one of which contains the interior of $X$ and the other, $X_C$, is disjoint from $X$. If the inclusion map $C\to \bar{X}_C$ (the closure of $X_C$ in $M$) does not induce a surjection $H_1(C)\to H_1(\bar{X}_C)\cong H_1(X_C)$, then applying the Mayer-Vietoris sequence, we again see that $H_1(X)\to H_1(M)$ is not surjective, which is a contradiction. In particular, by the classification of compact surfaces, either $\bar{X}_C$ is homeomorphic to a closed disk, or $X_C$ is noncompact and, thus, $H_1(C)\to H_1(X_C)$ is 1-1.
We, therefore, enlarge $X$ by adding to it the complementary components $X_C$ homeomorphic to $R^2$. The new subsurface with boundary, denoted $Z$, still has the property that the inclusion $Z\to M$ induces a surjective homomorphism $H_1(Z)\to H_1(M)$ (since $Z$ contains $X$). Now, however, the homomorphism $H_1(Z)\to H_1(M)$ is also injective. Therefore, $H_1(Z)\to H_1(M)$ is an isomorphism.
Let $N$ denote the interior of $Z$: It is a noncompact tame surface ($Z=\bar{N}$).
We still have isomorphisms induced by inclusions
$$
H_1(N)\cong H_1(Z)\cong H_1(M).
$$
So far, the proof was self-contained. I will now use:
Theorem 2. Suppose that $S'\subset S$ is a connected subsurface $S'$ in a noncompact connected surface $S$ such that the inclusion $S'\to S$ induces an isomorphism $H_1(S')\to H_1(S)$. Then $S'$ is homeomorphic to $S$.
See Theorem 7.1 in
Martin Goldman, An algebraic classification of noncompact 2-manifolds.
Trans. Amer. Math. Soc. 156 (1971), 241–258.
This theorem implies that the surfaces $N$ and $M$ are homeomorphic. Since $N$ is tame, so is $M$. qed
Remark. Since fundamental groups of noncompact connected surfaces are free, finiteness of $H_1(M; Z_2)$ is equivalent to the finite generation of $\pi_1(M)$. This allows for an alternative geometric proof of the implication (2)$\Rightarrow$(1) which I give below. (But the proof uses much heavier machinery.)
Equip $M$ with an arbitrary Riemannian metric. This, in turn, defines a conformal structure on $M$. Lift this conformal structure to the universal covering $\tilde{M}\to M$. By the Uniformization Theorem, $\tilde{M}$ is conformal to either the open unit disk or to the complex plane. Accordingly, we obtain a properly discontinuous, free, isometric action of $\Gamma=\pi_1(M)$ on the hyperbolic plane ${\mathbb H}^2$ (in the unit disk model) or the Euclidean plane $E^2$. I will consider the former case since the latter is easier. We have a discrete finitely generated subgroup $\Gamma< Isom({\mathbb H}^2)$. Every such subgroup admits a finitely-sided fundamental polygon $F\subset {\mathbb H}^2$, see for instance Alan Beardon's book "The Geometry of Discrete Groups."
The sides of $F$ are pairwise identified by some generators $\gamma_1,...,\gamma_n$ of $\Gamma$ and the quotient of $F$ by this identification is homeomorphic to our surface $M$. The polygon $F$ is noncompact (since $M$ is). Its noncompactness stems from two possible sources:
(a) $F$ may have "ideal vertices", points in the (ideal) boundary circle of ${\mathbb H}^2$, where two of the sides of $F$ meet.
(b) $F$ may have "ideal edges", arcs in the boundary of ${\mathbb H}^2$.
We enlarge $F$ by adding to it the ideal vertices and edges, the result is $\bar{F}$, a compact polygon. Identifying its boundary edges results in a compact surface with boundary $S$: The boundary of $S$ is the image of the union of ideal edges of $F$. The (finitely many) ideal vertices of $F$ project to finitely many points $p_1,...,p_k$ in $S- \partial S$. Thus, $M$ is homeomorphic to the complement
$$
S- (\partial S \cup \{p_1,...,p_k\}).
$$
Equivalently, let $D_1,...D_k$ denote small (pairwise disjoint and disjoint from $\partial S$) closed disk neighborhoods of $p_1,...,p_k$. Then $M$ is homeomorphic to the complement
$$
S- (\partial S \cup D_1\cup ...\cup D_k).
$$
Let $\bar{M}$ denote $M- (cl(D_1)\cup ...\cup cl(D_k))$. Then $\bar{M}$ is a compact surface with boundary and $M$ is homeomorphic to the interior of $M$. qed
Edit. Here is one more proof of the implication (3)$\Rightarrow$(1). It uses Lemma 2.2 in
D. B. A. Epstein, Curves on 2-manifolds and isotopies. Acta Math. 115 1966 83–107:
Lemma 1. Suppose that $S$ is a connected surface and $G$ is a finitely-generated subgroup of $\pi_1(S)$. Then there exists a compact $\pi_1$-injective subsurface $S'\subset S$ such that the image of $\pi_1(S')$ in $\pi_1(S)$ contains $G$.
Now, assuming that $\pi_1(S)$ is finitely-generated and $G=\pi_1(S)$, we obtain a compact subsurface $S'$ whose fundamental group is isomorphic to $\pi_1(S)$ via the map induced by the embedding $S\to S'$. Now one can either quote Goldman's result to conclude a homeomorphism $int(S')\to S$, or directly argue that each component of $S- int(S')$ is homeomorphic to $S^1\times [0,1)$:
Lemma 2. Suppose that $A$ is a connected noncompact surface with connected boundary $C=\partial A$, such that the inclusion $C\to A$ induces an isomorphism of fundamental groups. Then $A$ is homeomorphic to $S^1\times [0,1)$.
Proof. Consider the surface $B$ obtained by attaching a closed disk $D$ to $A$ along $C$. Then, by the van Kampen theorem, $B$ is simply-connected (since $\pi_1(C)\to \pi_1(A)$ is surjective). Therefore, since $B$ is noncompact, $B$ is homeomorphic to ${\mathbb R}^2$, see e.g. Corollary 1.8 in Epstein's paper. By the 2-dimensional Schoenflies theorem, there is a homeomorphism ${\mathbb R}^2\to {\mathbb R}^2$ sending $D$ to the closed unit disk $D(0,1)$ centered at the origin. This homeomorphism sends $A$ to the complement to the interior of $D(0,1)$, which is homeomorphic to $S^1\times [0,1)$. qed