The answer is negative. Suppose for contradiction that $S$ is such a surface.
Fix $g\geq 24$. Then the coarse moduli space of curves $M_g$ is not of general type, 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 $H\to M_g$ to the coarse moduli space of genus $g$ curves 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.
Now let $H_i\to Pic(S)$ be the natural morphism. A general non-empty fiber of it is an open subset of a linear system on $S$, hence is covered by (open subsets of) rational curves. By our choice of $g$, the very general such fiber is contracted by $H_i\to M_g$. Consequently, $H_i\to Pic(S)$ cannot be constant. This 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 $S$.