The answer is no.

I find it conceptually easier to work in the category of hyperbolic surfaces (but these are Riemann surfaces too, so it is permitted!).

We build $X$ by gluing together an infinite collection of surfaces $X_k$ (for $k \in \mathbb{Z}$) with each $X_k$ a copy of the genus one surface with two boundary components. The boundary components all have length one (say) and $X_k$ glues to $X_{k \pm 1}$. We arrange matters so that there is a simple closed hyperbolic geodesic in $X_k$ with length less than $1/|k|$. It follows that $X$ does not cover any closed surface.

Edit: as Moishe points out, my answer (just below) is for a different, and easier, question. I will leave the answer up, as it does feel “related”.

The answer is no.

I find it conceptually easier to work in the category of hyperbolic surfaces (but these are Riemann surfaces too, so it is permitted!).

We build $X$ by gluing together an infinite collection of surfaces $X_k$ (for $k \in \mathbb{Z}$) with each $X_k$ a copy of the genus one surface with two boundary components. The boundary components all have length one (say) and $X_k$ glues to $X_{k \pm 1}$. We arrange matters so that there is a simple closed hyperbolic geodesic in $X_k$ with length less than $1/|k|$. It follows that $X$ does not cover any closed surface.