The answer is negative. Suppose for contradiction that $S$ is such a surface.

 Fix $g\geq 24$. Then the coarse moduli space of genus $g$ curves $M_g$ is of general type (this is due to Harris, Mumford and Eisenbud, see for instance [The Kodaira dimension of the moduli space of curves of genus $\geq 23$]), hence a fortiori not uniruled.
 Let $H$ be the Hilbert scheme of smooth curves of genus $g$ in $S$ and let $(H_i)_{i\in I}$ be its irreducible components: the index set $I$ is countable. By hypothesis, the classifying morphism $H\to M_g$ is surjective at the level of $\mathbb{C}$-points. By a Baire category argument, there exists $i\in I$ such that $H_i\to M_g$ is dominant. 

Suppose that the natural morphism $H_i\to Pic(S)$ is constant. Then $H_i$ is an open subset of a linear system on $S$, hence is covered by (open subsets of) rational curves. By our choice of $g$, $H_i\to M_g$ cannot be dominant, which is a contradiction. Thus, $H_i\to Pic(S)$ cannot be constant, which proves that $Pic^0(S)$ cannot be trivial. Equivalently, the Albanese variety $A$ of $S$ is not trivial.

Consider the Albanese morphism $a:S\to A$. The curves embedded in $S$ are either contracted by $a$ or have a non-trivial morphism to $A$. Those that are contracted by $a$ form a bounded family, hence have bounded genus. Curves $C$ that have a non-trivial morphism to $A$ are such that there is a non-trivial morphism $Jac(C)\to A$, but this is impossible if $Jac(C)$ is simple of dimension $>\dim(A)$. Consequently, a smooth curve with simple jacobian that has high enough genus cannot be embedded in $S$. This concludes because a very general curve of genus $g$ has simple jacobian (there even exist hyperelliptic such curves by [Zarhin, Hyperelliptic jacobians without complex multiplication]).