$\newcommand{\fs}{\mathscr{F}}\newcommand{\gs}{\mathscr{G}}$Let ${\bf C}$ be some algebraic category (e.g. commutative rings), ${\bf Psh}(X,{\bf C})$ the category of presheaves on $X$ taking values in ${\bf C}$, and ${\bf Sh}(X,{\bf C})$ the category of sheaves on $X$ taking values in ${\bf C}$. Let $\phi:\fs\to\gs$ be an epimorphism in ${\bf Sh}(X,{\bf C})$. Is it possible that $\phi$ is not an epimorphism in ${\bf Psh}(X,{\bf C})$ ?
(Sheaf monomorphism seems to imply presheaf monomorphism, if one assumes that every presheaf can be sheafified. The proof doesn't extend to epimorphisms.)
No. If sheaf epimorphisms were always presheaf epimorphisms then there would be no point to sheaf cohomology!
Here's an example. Let $X$ be $\mathbb{R}^2$ minus a point, and let $\Omega^*_X$ be the de Rham complex. $X$ is two dimensional, so we get an exact sequence $$\Omega^0_X \longrightarrow \Omega^1_X \longrightarrow \Omega^2_X \longrightarrow 0$$ in $\textbf{Sh}(X, \textbf{Vect}_\mathbb{R})$, because we can always locally integrate a closed $(n + 1)$-form to get an $n$-form using the Poincaré lemma. Let $B^1 (\Omega_X)$ be the image of $\Omega^0_X \to \Omega^1_X$ in $\textbf{Sh}(X, \textbf{Vect}_\mathbb{R})$. As usual, $\Omega^0_X \to B^1 (\Omega_X)$ is an epimorphism, but after taking global sections, $\Gamma (X, \Omega^0_X) \to \Gamma (X, B^1 (\Omega_X))$ is not. Indeed, we find that the $1$-form $$\mathrm{d} \theta = \frac{-y \, \mathrm{d} x + x \, \mathrm{d} y}{x^2 + y^2}$$ is closed, but $\mathrm{d} \theta$ cannot be exact because $$\int_S \mathrm{d} \theta = 2 \pi$$ where $S$ is the unit circle in $X$. Thus $\Gamma (X, \Omega^0_X) \to \Gamma (X, B^1 (\Omega_X))$ is not surjective, which is just as well, because de Rham's theorem tells us that $H^1(\Gamma(X, \Omega^*_X)) \cong H^1(X, \mathbb{R}) \cong \mathbb{R}$!