I claim that, if $\mathcal{D}$ admits arbitrary coproducts, then for any $c \in \mathcal{C}, d \in \mathcal{D}$, there is an object $h_{c,d} \in \mathrm{Fun}(\mathcal{C}, \mathcal{D})$ such that $\hom(h_{c,d}, G) = \hom(d, Gc)$. Given such an object $h_{c,d}$, it follows that a monomorphism of functors $G \to G'$ must have the property that $\hom(d, Gc) \to \hom(d, G'c)$ is injective for each pair of objects $d, c$, meaning that you have a pointwise monomorphism (i.e. $Gc \to G'c$ is a monomorphism). To get such a functor, you imitate the Yoneda lemma: let $h_{c,d}(c') = \hom(c, c') \times d$ (since $\mathcal{D}$ admits arbitrary coproducts, it is tensored over sets and the preceding makes sense). Now given an arbitrary functor $G$ and given $d \to Gc$, one extends to $h_{c,d}(\cdot) = \hom(c,
\cdot) \times d \to G$ in the natural way. I won't write out the details, but if I'm not being silly then what I claimed is true.
Now to the question on epimorphisms. Ideally, I want to claim that there is a functor $h'_{c,d}$ such that mapping into it from another functor $G$ is the same as giving a map $Gc \to d$. In other words, if we take the functor from $\mathrm{Fun}(\mathcal{C}, \mathcal{D})$ to $\mathcal{D}$ given by "evaluation" at $c$, then there is a right adjoint to this evaluation (previously the $h_{c,d}$ construction showed that there was a left adjoint). If there is such a right adjoint, then we're done.
In fact, let's state the whole argument more categorically:
A left adjoint preserves epimorphisms. A right adjoint preserves monomorphisms.
This is not hard at all (if $(F, G)$ is an adjunction and $c \to c'$ is an epimorphism, then we have to show that $Fc \to Fc'$ is one too; that is, $\hom(Fc', d) \to \hom(Fc, d)$ is always an injection of sets. But this is $\hom(c,' Gd) \to \hom(c, Gd)$).
So the point of the above discussion is that if $\mathcal{D}$ admits coproducts, then the evaluation functor at $c$, $\mathrm{Fun}(\mathcal{C}, \mathcal{D}) \to \mathcal{D}$ is a right adjoint. Thus it sends "global" monomorphisms of functors to monomorphisms by evaluation at each $c \in \mathcal{C}$, and thus "global" monomorphisms are "pointwise" epimorphisms. (The converse, as you observed, is easy.) Now we can try to the same thing for epimorphisms. As above, the main point is to construct a right adjoint to the evaluation functor. This can be done whenever $\mathcal{D}$ has all products. Namely, given $c ,d$, we consider the functor $c \mapsto d^{\hom(c, c')}$. One can check that this has the analogous universal property to $h_{c,d}$ above, and that it defines a left adjoint to the evaluation functor. So, if the category $\mathcal{D}$ has arbitrary products, then epimorphisms of natural transformations are pointwise epimorphisms.
One can ask the more general question: if $F: \mathcal{C} \to \mathcal{C}'$ is a functor, when is $F_*: \mathrm{Fun}(\mathcal{C}', \mathcal{D}) \to \mathrm{Fun}(\mathcal{C}, \mathcal{D})$ a left or right adjoint? The answer is that if $\mathcal{D}$ is cocomplete and complete, it's both. Since the "lower star" preserves limits and colimits (which are calculated pointwise, after all). Some specific examples are the lower star and upper star from pre(!)sheaf theory, and the skeleton and coskeleton functors on simplicial sets.
I'm not sure what the answer to your question is without such assumptions on $\mathcal{D}$.
Eric's answer is very nice, but addresses the case when $\mathcal{F}$ and $\mathcal{G}$ are sheaves of abelian groups, whereas the original question asks about sheaves of sets. I thought I might show how to modify things in this case. The idea is quite similar; one again wants to use a skyscraper sheaf (as noted in the OP) to make an ''indicator function''-like argument. I'll use the notation of Vakil's ''Foundations of Algebraic Geometry''.
Let $\phi \colon \cal{F} \to \cal{G}$ be our epimorphism of sheaves. Define $\rho \colon \cal{G} \to i_{p, \ast}\{0, 1\}$ by
$$\rho(U)(g) = \begin{cases} 0 \text{ if } (g, U) \in Im(\phi_{p}) \\
1 \text{ otherwise } \end{cases}$$
for open sets $U$ containing $p$, and trivial maps otherwise. It is straightforward to see that $\rho$ is a morphism of sheaves. Now define $\alpha \colon \cal{G} \to i_{p, \ast}\{0, 1\}$ by
$$\alpha(U)(g) = 0$$
for open sets $U$ containing $p$, and trivial maps otherwise. Likewise, it is obvious that $\alpha$ is a morphism of sheaves. Furthermore, it is clear that $\rho \circ \phi = \alpha \circ \phi$, so $\rho = \alpha$, since $\phi$ is an epimorphism. Surjectivity of the stalk maps is is now easily deduced.
Best Answer
Let $X$ be a topological space, and let $\alpha : \mathscr{F} \to \mathscr{G}$ be a morphism of sheaves on $X$. Let $j : \{ x \} \hookrightarrow X$ be the inclusion of a point into $X$. Then, by definition, the stalk of $\mathscr{F}$ at $x$ is just the inverse image sheaf $j^* \mathscr{F}$. But we know $j^* \dashv j_*$ (i.e. the inverse image functor $j^*$ is left adjoint to the direct image functor $j_*$), and left adjoints preserve colimits, hence if $\alpha$ is epic, so is $\alpha_x : \mathscr{F}_x \to \mathscr{G}_x$.
Conversely, suppose $\alpha_x : \mathscr{F}_x \to \mathscr{G}_x$ is surjective for every $x$ in $X$. Let $\beta, \gamma : \mathscr{G} \to \mathscr{H}$ be two further morphisms of sheaves, and suppose $\beta \circ \alpha = \gamma \circ \alpha$. We need to show $\beta = \gamma$. Certainly, we have $\beta_x \circ \alpha_x = \gamma_x \circ \alpha_x$, and $\alpha_x$ is epic by hypothesis, so we have $\beta_x = \gamma_x$ for all $x$ in $X$. Let $s \in \mathscr{G}(U)$, and consider $\beta_U(s)$ and $\gamma_U(s)$. Since $\beta$ and $\gamma$ agree on germs, at each $x$ in $U$ there is an open neighbourhood $V$ on which $\beta_V(s|_V) = \gamma_V(s|_V)$, and so by the unique collation property, we must have $\beta_U(s) = \gamma_U(s)$, and therefore we indeed have $\beta = \gamma$.