are two definitions of a *minimal* chain complex equivalent? In Eisenbud, a complex $C_\ast$ over a local ring $(R,\mathfrak{m})$ is minimal when $C_\ast\!\otimes_R\!\frac{R}{\mathfrak{m}}$ has zero boundaries.
My more intuitive definition, for a chain complex of free modules over any ring $R$, is: Given homotopy equivalent chain complexes of free modules $C_\ast=R^{(I_\ast)}$ and $C'_\ast=R^{(I'_\ast)}$, let $\leq$ be a componentwise partial order given by $C_\ast\leq C'_\ast$ iff $\forall k\!: |I_k|\leq |I'_k|$. Then $C_\ast$ is minimal when it is a minimal element among all chain complexes of free modules which are homotopy equivalent to $C_\ast$. In other words, $C_\ast$ is minimal when for any $C'_\ast\simeq C_\ast$ with $\forall k\!:|I'_k|\leq |I_k|$ we have $C'_\ast\!\cong\!C_\ast$.

For chain complexes of free modules over a local ring, are the two
  definitions equivalent?

The underlying problem here: do h-equivalent complexes with equal ranks need to be isomorphic?

Are there any sufficient conditions, under which a minimal complex is
  the smallest complex?

If $R$ is a PID and all modules $C_k$ are finitely-generated, then a minimal complex always exists, and is the smallest complex, obtained from the Smith normal forms of all $\partial_k$ by cancelling out all $1$s.
But what about over large rings, such as $A\otimes A^{op}$ where $A=\Lambda[x_1,\ldots,x_n]$ or over $K[x_1,\ldots,x_n|f_1,\ldots,f_r]$ or $K\langle x_1,\ldots,x_n|f_1,\ldots,f_r\rangle$?
 A: I'm not sure whether you're mainly thinking of complexes of finitely generated free modules? If not, then your definition seems rather unnatural, as you could have a complex $C_*$ of infinitely generated free modules where taking the direct sum of $C_*$ with a non-zero contractible complex didn't change the rank of any $C_n$.
So in what follows I'll assume we're only considering complexes of finitely generated free modules.
With that restriction, the answer to your first question is yes:
Eisenbud's definition is equivalent to the condition that $C_*$ has no non-zero contractible direct summand or, equivalently, no direct summand of the form $\dots\to0\to R\stackrel{1}\to R\to0\to\dots$. For suppose that some differential $d_n:C_n\to C_{n-1}$ does not have the property that $d_n\otimes \frac{R}{\mathfrak{m}}=0$. Then there is a projection $\alpha:C_{n-1}\to R$ such that $\alpha d_n$ is surjective, giving a surjective chain map from $C_*$ to a complex of the form $\dots\to0\to R\stackrel{1}\to R\to0\to\dots$, which it is easy to see must split, since $\alpha d_n$ splits. The other implication is trivial.
Now it follows easily that a complex (of finitely generated free modules over a local ring) that is minimal in your sense is minimal in Eisenbud's, as otherwise you could get a smaller homotopy equivalent complex by removing a contractible direct summand.
Now suppose $C_*$ is minimal in Eisenbud's sense and that $C_*'$ is another homotopy equivalent complex with $C_*'\leq C_*$. Let $f:C_*\to C_*'$ and $g:C_*'\to C_*$ be a homotopy inverse pair of maps, so that $gf\simeq\operatorname{id}_{C_*}$. Since $\operatorname{id}_{C_*}-gf=hd+dh$ for some chain homotopy $h$ and $d\otimes_R\frac{R}{\mathfrak{m}}=0$, $gf\otimes_R\frac{R}{\mathfrak{m}}$ is the identity map on $C_*\otimes\frac{R}{\mathfrak{m}}$, and so $gf$ is an isomorphism. Thus $C_*$ is a direct summand of $C_*'$, contradicting $C_*'\leq C_*$ unless $C_*'\cong C_*$.
I don't know a good answer to your second question, but here are a few observations:
First, for a ring without the Invariant Basis Number property your definition seems a bit unnatural, for much the same reason it does for infinitely generated modules.
Secondly, it also seems a bit unnatural if your ring $R$ has projective modules that are not free. Consider, for example, $R=\mathbb{Q}\times\mathbb{Q}$, for which $R=P\oplus Q$ is a direct sum of two non-isomorphic non-zero projective modules, and consider the complex 
$$\dots\to0\to0\to P\to0\to0\to\dots$$
(with $P$ in degree zero). Of course, this is not a complex of free modules, but you could get two different (unbounded) homotopy equivalent complexes of free modules that are both minimal in your sense, by taking the direct sum with a contractible complex of the form
$$\dots\to\to0\to Q\to R\to R\to\dots$$
or of the form
$$\dots\to R\to R\to Q\to0\to0\to\dots$$
(with $Q$ in degree zero in both cases).
I guess my point is that, for a ring where not all projectives are free, it seems rather unnatural to look for "minimal" complexes of free modules. After all, the original complex
$$\dots\to0\to0\to P\to0\to0\to\dots$$
seems "minimal" in any reasonable sense!
Finally, another possible definition, that makes sense for any ring (even for complexes of infinitely generated modules, and that agrees with Eisenbud's for complexes of free modules over local rings, is that a complex is "minimal" if it has no non-zero contractible direct summands.
