As you may possibly already be aware, there is a parallel phenomenon in circle packing riemann surfaces.
Those compact riemann surfaces admitting full circle packings are a countable dense subset of the moduli space, where by full circle packing a riemann surface, I mean finding a circle packing $C$ such that the carrier of the nerve graph $T_C$ coincides with our surface. This is due to R. Brooks, "Circle packings and co-compact extensions of Kleinian groups", Invent. Math. 86, 1986, 461-469.
But G. Brock Williams has also proven that all noncompact riemann surfaces are packable:
"Noncompact surfaces are packable", J. D'Analyse Math, Vol 90, 2003, 243-255.
The first basic 'intuition' as to why noncompact riemann surfaces are packable is of course that any obstructions to completing a (partial) circle packing to a full packing can be `hidden' into the cusp (similar to how any noncompact surface admits lots of nonvanishing vector fields).
I would be very interested in reading your proof that noncompact riemann surfaces admit Belyi functions. In particular, I would like to know if your functions have the same "flat euclidean" structure up within the cusps as Williams' paper. Specifically, the final corollary 6.1 in Williams states "every noncompact riemann surface of finite type supports a circle packing asymptotic to the euclidean ball-bearing packing in the cusps". I have to admit that I do not find Williams' paper to be entirely educating, and am very anxious to see other approaches.
The answer is negative. Suppose for contradiction that $S$ is such a surface, and let me first assume that it is smooth and projective.
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 $H_i\to \operatorname{Pic}(S)$ 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, $H_i\to \operatorname{Pic}(S)$ cannot be constant, which proves that $\operatorname{Pic}^0(S)$ 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 $\operatorname{Jac}(C)\to A$, but this is impossible if $\operatorname{Jac}(C)$ 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]).
As pointed out in the comments, the argument needs to be modified if $S$ is non-algebraic or singular. I explain now these additional arguments.
If $S$ is smooth but non-algebraic, we can use the following (probably overkill) variant. We can consider the open space $H$ of the Douady space of $S$ parametrizing smooth connected curves in $S$. It is a countable union of quasiprojective varieties by [Fujiki, Countability of the Douady space of a complex space] and [Fujiki, Projectivity of the space of divisors on a normal compact complex space]. It has an analytic morphism to the analytic space $\operatorname{Pic}(S)$ parametrizing line bundles on $S$ [Grothendieck, Techniques de construction en géométrie analytique IX §3]. The dimension of $\operatorname{Pic}(S)$ is finite equal to $h^1(S,\mathcal{O}_S)$. The dimension of the fibers of $H\to\operatorname{Pic}(S)$ is at most $1$. Indeed, otherwise, we would have a linear system of dimension $>1$ on $S$ consisting generically of smooth connected curves, hence a dominant rational map $S\dashrightarrow\mathbb{P}^2$, showing that $S$ is of algebraic dimension $2$, hence that $S$ is algebraic. It follows that every connected component of $H$ has dimension $\leq h^1(S,\mathcal{O}_S)+1$. Now choose $g$ such that $\dim(M_g)>h^1(S,\mathcal{O}_S)+1$, and let $H_g$ be the union of connected components of $H$ parametrizing genus $g$ curves. A Baire category argument applied to the image of $H_g\to M_g$ shows that there are genus $g$ curves that do not embed in $S$, as wanted.
Finally, if $S$ is singular, consider a desingularization $\tilde{S}\to S$. The hypothesis on S implies that every smooth projective curve may be embedded in $\tilde{S}$, with the exception of at most a finite number of isomorphism classes of curves (namely, the curves that are connected components of the locus over which $\tilde{S}\to S$ is not an isomorphism). The arguments above apply as well in this situation.
Best Answer
You should probably check the following article: Migliorini, Luca, "On the compactification of Riemann surfaces". Here is the Mathscinet review about it: "In this paper the author studies some questions concerning the compactifications of Riemann surfaces. It is proved that if $X$ is an open connected Riemann surface then X has finite genus if and only if there exists a holomorphic injection $i: X \hookrightarrow \tilde{X}$ (with $\tilde{X}$ a compact Riemann surface), $i(X)$ being dense in $\tilde{X}$..."