A *finite* first order theory whose finite models are exactly the $\Bbb F_p$? Since this question turned out to be trivial, I'm now asking this strengthened version: 
Is there a finitely axiomatized first order theory $T$ in the language of rings such that its finite models are exactly the fields $\Bbb F_p$ with $p$ prime (but no $\Bbb F_q$ with $q$ a proper prime power is a model of $T$)?
 A: EDIT:  This answer is only a proof of a special case of
the given question (the case when the infinite models of the
given theory are exactly the infinite fields).  It does not
solve the general question.
Theorem.  Let $\mathcal{M}$ be the class of all fields $F$ such
that either $F$ is infinite or $F$ is isomorphic to $\Bbb{F}_p$ for 
some prime $p$.  Then $\mathcal{M}$ has no finite FOL-axiomatization.
Proof.  For any prime $p$ and any integer $n \geq p$, let
$\sigma_{p,n}$ be the translation of the following statement of
ring theory into a FOL-sentence:

If $R$ has cardinality $n$ and $p \cdot 1_R = 0_R$, then for
  every ring element x, x is either $0_R$, $1_R$, $2 \cdot 1_R$, 
  $\ldots$, or $(p-1) \cdot 1_R$.

Let $\Sigma$ be the union of the first-order field axioms and the
set of all FOL-sentences $\sigma_{p,n}$ for some prime number $p$
and for some integer $n \geq p$.  Then the class of all models of
$\Sigma$ is clearly $\mathcal{M}$.
Suppose now that the class $\mathcal{M}$ has a finite FOL-axiomatization
$\{ \rho_1, \ldots, \rho_m \}$.  Let $\rho$ be the conjunction of
the sentences $\rho_i$ for $i = 1, \ldots, m$.  Then since $\Sigma$
is also an axiomatization of $\mathcal{M}$, $\rho$ follows logically
from $\Sigma$.  Thus, $\rho$ follows logically from a finite subset
$\Sigma_0$ of $\Sigma$ by the compactness theorem of first-order
logic.  Then $\Sigma_0$ is a finite axiomatization of the class
$\mathcal{M}$.
Since $\Sigma_0$ is finite and there are infinitely many primes,
we can choose a prime $\hat{p}$ and an integer $\hat{n} \geq \hat{p}$
such that for every prime $p \geq \hat{p}$ and for every $n \geq
\hat{n}$, $\sigma_{p,n}$ is not an element of $\Sigma_0$.  Then,
clearly, every field $\Bbb{F}_{p^n}$ for $p \geq \hat{p}$ and $n
\geq \hat{n}$ is a model of $\Sigma_0$ and must therefore be an
element of $\mathcal{M}$.  But this contradicts our initial assumption
for $\mathcal{M}$.  Therefore $\mathcal{M}$ has no finite
FOL-axiomatization.
Q.E.D.
A: I believe the answer is no. Ax's classical paper The Elementary Theory of Finite Fields provides a (nice) algorithm for deciding the set of finite fields satisfying a given statement $E$. A few paragraphs after the statement of the main theorem there is the following remark:

If $E$ holds for all prime fields, then $E$ holds for all finite fields of sufficiently large characteristic.

which gives a negative answer to your question.
