Since the finite group must act on $X$ discontinuously, the quotient space $X/G$ should be an orbifold, and it seems that the 2 dimensional orbifolds are all homeomorphic to manifolds with boundaries. If the group action does not preserve orientation of the surface, like a reflection, then a boundary will occur in the quotient space $X/G$. Will this also happen in the orientation preserving case?
I tried to show that a finite order orientation preserving automorphism only have finitely many fixed points, but I do not know whether there is counter example. If this is true, then every fixed point will result in a cone point rather than a boundary point in the orbifold $X/G$.
If you are working in the smooth category, the result is easy (if you know some basic differential geometry). In the topological category, this is actually a nontrivial theorem, essentially due to Kerékjártó. You can find a modern proof of the Kerékjártó theorem in
Constantin, Adrian; Kolev, Boris, The theorem of Kerékjártó on periodic homeomorphisms of the disc and the sphere, Enseign. Math., II. Sér. 40, No. 3-4, 193-204 (1994). ZBL0852.57012.
The paper is freely available on arXiv here. The reduction from a theorem about periodic homeomorphisms of the sphere and the plane to the case of surfaces takes a bit of thought, you can use, for instance, Proposition 2.4 in the linked paper and the main theorem to prove:
Suppose that $S$ is a topological surface (compactness is irrelevant), $f: S\to S$ is a periodic homeomorphism and $x\in S$ is a fixed point of $f$. Then there is an open disk neighborhood $U$ of $x$ invariant under $f$ such that the restriction of $f$ to $U$ is topologically conjugate to a rotation or a reflection acting on the plane.