I don't think so. Let $X$ be $\mathbb A^1_{\mathbb C}$ with two points glued together, and let $Y$ be the standard double étale cover, obtained by identifying two copies of $\mathbb A^1$. I claim that the étale sheaf defined by $Y$ does not come from the Zariski topology. This follows from the fact the stalk at the node of the restriction of the sheaf to the Zariski topology is empty, while the stalk of the étale sheaf consists of two points.
The following argument shows that if $C$ is essentially large (and locally small), then the presheaf $\Omega$ of sieves on $C$ is not small. Unfortunately, as Zhen points out, this does not show that $\mathcal{P}C$ does not have a subobject classifier, since its subobject classifier ought to be the presheaf of small sieves (i.e., sieves such that the corresponding presheaf is a small presheaf). In general, not all sieves are small, and in fact it seems rather difficult to construct any small sieves (exercise: construct a locally small category in which every object has a large set of sieves, but not a single nontrivial small sieve). It thus seems unlikely to me that this argument can be turned into a proof that $\mathcal{P}C$ does not have a subobject classifier without some strong additional hypotheses. Nevertheless, I'm keeping it as an answer in case you may find it somehow helpful (and because it is a nifty argument, even thought it doesn't prove what I hoped it would!).
Suppose that $\Omega$ is a small presheaf. Then every object $a\in C$ has only a small set of sieves, and there is a small set $S$ of objects such that for every $a\in C$ and every sieve $\sigma\in\Omega(a)$, there is an object $s\in S$, a map $f:a\to s$, and a sieve $\tau\in\Omega(s)$ such that $\sigma=f^*\tau$. In fact, if any such $\tau$ exists, then there is a canonical choice, namely $\tau=f_*\sigma$.
For any $a\in C$, let $\sigma_a$ be the sieve consisting of all maps $b\to a$ that do not have a right inverse (this is the second-largest sieve on $a$). We can thus find some $s_a\in S$ and $f_a:a\to s_a$ such that $\sigma_a=f_a^*f_{a*}\sigma_a$. Let's figure out explicitly what the condition $\sigma_a=f_a^*f_{a*}\sigma_a$ means. The sieve $f_{a*}\sigma_a$ consist of all maps of the form $f_ag$ where $g:b\to a$ does not have a right inverse. The sieve $f_a^*f_{a*}\sigma_a$ then consists of all maps $h$ such that $f_ah=f_ag$ for some $g$ that does not have a right inverse. The equation $\sigma_a=f_a^*f_{a*}\sigma_a$ holds iff the latter sieve does not contain the identity $1:a\to a$, or equivalently if every $g:a\to a$ such that $f_a=f_ag$ has a right inverse.
Since $S$ is small and $C$ is essentially large, we can find an essentially large set of objects $D\subseteq C$ such that all the objects $a\in D$ have the same associated object $s_a\in S$. That is, there is a single object $s\in S$ such that for each $a\in D$, there is a map $f_a:a\to s$ such that if $g:a\to a$ is such that $f_a=f_ag$, then $g$ has a right inverse. Now consider the sieves on $s$ generated by these maps $f_a$. Since $\Omega(s)$ is a small set, there is an essentially large set of objects $E\subseteq D$ such that the $f_a$ for $a\in E$ all generate the same sieve on $s$. That is, for any $a,b\in E$, there exist maps $g:a\to b$ and $h:b\to a$ such that $f_bg=f_a$ and $f_ah=f_b$. But then $f_ahg=f_a$, so $hg$ has a right inverse, and similarly $gh$ has a right inverse. In particular, $g$ and $h$ both have right inverses, and so $a$ and $b$ are retracts of each other. But for any object $a$, there are only an essentially small set of objects that are retracts of $a$, and this contradicts the essential largeness of $E$.
Best Answer
The good setting to answer this question really is that of algebraic spaces. Then I believe that the answer is no, basically because every sheaf $F$ on the small étale site is representable by an étale $X$-algebraic space. The algebraic space that represents $F$ is constructed as follows: consider the collection of pairs $(V,x)$ with $V$ an affine étale $X$-scheme and $x\in F(V)$. Let $U$ be the disjoint sum of the $V$'s indexed by this collection. If $(V,x)$ and $(V',x')$ are two pairs, the natural morphism $V\times_FV'\to V\times_XV'$ is an open immersion. This shows that $U\times_FU$ is open in $ U\times_X U$ and hence defines an étale equivalence relation on $U$. The $X$-algebraic space we're looking for is the quotient of this equivalence relation.
This is due to Michael Artin, reference: his book "Théorèmes de représentabilité pour les espaces algébriques", chap. VII. Le théorème de finitude en cohomologie étale, §1. Définition des faisceaux constructibles.