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 $E$.
It's true, although maybe somewhat tedious to prove directly from scratch. I'll prove some intermediate results first (also for easy reference). I claim no originality, but couldn't find a reference anywhere.
We denote by $1_A : X\to 2$ the indicator function of $A\subseteq X$.
Lemma: In $\mathsf{Meas}$ (the category of measurable spaces and maps) a morphism $e : S\to E$ is epi iff it's surjective.
Proof: The forgetful functor into $\mathsf{Set}$ is faithful, hence surjective implies epi. Conversely, let $e$ be epi. Since
$$1_{e(S)} \circ e = 1 = 1\circ e,$$
and any map into $2$ is automatically measurable,
we have $1_{e(S)} = 1$, i.e. $e$ is surjective. $\square$
Lemma: In $\mathsf{Meas}$ a morphism $i : S \to X$ is a strong mono iff it's a subspace embedding, i.e. $i$ is injective and the $\sigma$-algebra $\mathcal F_S$ on $S$ is exactly the $\sigma$-algebra generated by $i$
$$\mathcal F_S = \sigma(i) = \{i^{-1}(A) : A\in \mathcal F_X\}.$$
Proof: Since $\mathsf{Meas}$ is topologically concrete is (bi-) complete, hence extremal = strong. Hence, we can work with extremal monos.
"$\Leftarrow$":
Suppose $i : S\to X$ is a subspace embedding, which factors as $i = \phi \circ e$ in $\mathsf{Meas}$ with $e$ an epi. We need to show $e$ is iso.
Since $i$ is injective, so is $e$ and hence $e$ is bijective. We need to show that $e^{-1}$ is measurable, i.e. we claim
$$e(i^{-1}(A)) = (e^{-1})^{-1}(i^{-1}(A)) \in \mathcal F_E$$
for any $A\in \mathcal F_X$. Indeed, $e(i^{-1}(A)) = e \circ e^{-1}(\phi^{-1}(A)) = \phi^{-1}(A)\in \mathcal F_E$, since $\phi$ is measurable. Thus, $e$ is an iso and in conclusion $i$ is an extremal mono.
"$\Rightarrow$": Suppose $i$ is an extremal mono. Then $e : S\to i(S), x\mapsto i(x)$ is a well-defined surjection. We consider $i(S)$ as the image of $i$ equipped with subspace $\sigma$-algebra
$$\mathcal F_{i(S)} = \{i(S) \cap A : A\in \mathcal F_X\}.$$
Then $e^{-1}(i(S)\cap A) = i^{-1}(i(S)\cap A)\in \mathcal F_S$, so $e$ is measurable and hence an epi in $\mathsf{Meas}$.
Further, $i = e\circ \phi$, where $\phi : i(S)\to X, x\mapsto x$ is (measurable) the subspace inclusion. By assumption $e$ is iso. In particular,
$$\mathcal F_S = \sigma(e) = \{i^{-1}(A\cap i(S)) : A\in \mathcal F_X\} = \sigma(i),$$
where we used th injectivity of $i$ in the last step. Hence, $i$ is subspace embedding. $\square$
Proposition: $2$ is a strong subobject classifier in $\mathsf{Meas}$.
Proof: It's enough to show that it classifies subspace embeddings. Let $i : S \to X$ be a subspace embedding and $\top : 1\to 2$ be the map which maps the unique element in the singleton space / terminal object to $1 \in 2$.
Consider the diagram
The square commutes, since
$$1_{i(S)} \circ i = 1 = \top \circ !.$$
Now given $q$, such that
$$1_{i(S)} \circ q = 1$$
we have $q(Q) \subseteq i(S)$. Hence, for every $x\in Q$ there exists a $y\in M$, such that $i(y) = q(x)$. It mus be unique, since $i$ is injective. Hence, the map
$$u : Q\to M, x\mapsto y\text{ s.t. }i(y) = q(x)$$
is well-defined and satisfies $i\circ u = q$. Further, if there is another $u'$ with this property, then $i(u(x)) = q(x) = i(u'(x))$ for all $x\in Q$, hence $u = u'$ because $i$ is injective.
Therefore, the square is a pullback. We conclude that $2$ is classifier of strong subobjects. $\square$
Best Answer
If any of these categories had a subobject classifier, every monomorphism would be regular, so that's not happening.
The indiscrete two-point space is a strong subobject classifier in $\mathsf{Top}.$ Similarly, the subobject classifier for $\mathsf{Set}^{D^{\mathrm{op}}},$ equipped with the indiscrete topology on its values, is a strong subobject classifier for $\mathsf{Top}^{D^{\mathrm{op}}}.$ This is because the forgetful functor to presheaves induces an isomorphism between strong subobjects of a topological presheaf and arbitrary subobjects of a set-valued presheaf.