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The question is quite that what is in the title:

If $R$ and $S$ are homeomorphic Riemann surfaces, is it true that always exists a homeomorphism $h:R\to S$ which is orientation-preserving (at least in the case in which $R$ and $S$ are closed Riemann surfaces with genus $g$)?

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    Just to clarify, the fact that you have _Riemann_ surfaces (i.e., equipped with a holomorphic structure) is irrelevant to your question, right? – 2017-01-28
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    @AndrewD.Hwang Yes, I think so. I've said Riemann surfaces just because it is so in the problem I am studying, but I think the question can be made for orientable homeomorphic surfaces... – 2017-01-28
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    Thank you. My underlying motivation for asking was that "almost always" there is no _non-constant holomorphic map_ between two compact Riemann surfaces of the same genus, so the answer to your question depends on whether (or, as it happens, not) you want a _holomorphic_ homeomorphism. – 2017-01-28
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    The answer appears to be positive for noncompact surfaces as well, but you have to check carefully the classification of such surfaces say in http://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf – 2017-01-28

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Notice that this is equivalent to the question:

does every surface admit a orientation reverising self-homeo?

And the answer to this is yes. All compact orientable surfaces admit an embedding into $\mathbb R^3$ which is symmetric with respect to the $xy$ plane, and you can consider the reflection on that plane.

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    Thank you, @Mariano Suárez-Álvarez ! – 2017-01-28
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    Is it not true for non-compact orientable surfaces though? – 2017-01-28
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Here is an answer in the harder case, when you do not assume compactness of the surface. As Mariano noted in his answer, it suffices to show existence of an orientation reversing homeomorphism of every orieted surface.

First of all, clearly, it suffices to consider the case of connected surfaces. Such surfaces (without boundary!) are classified. In the orientable case (and you assume orientability) the complete set of topological invariants of such a surface $S$ is the following theorem which Richards attributes to Kerekjarto although Kerekjarto's proof had some gaps (see e.g. here for a detailed proof):

  1. The genus of $S$ (which can be infinite).

  2. The set of ends $e(S)$ equipped with a partition $e(S)= e_0(S) \sqcup e_1(S)$, where $e_0$ consists of the ends of genus $0$ (it is open in $e(S)$) and $e_1$ consists of ends of infinite genus. Elements of $e_0(S)$ can be characterized by the property that they admit planar neighborhoods in $S$.

The set of ends is a compact metrizable totally disconnected set. Therefore, it embeds in the real line $\subset {\mathbb C}\cup \{\infty\}$. Thus, every connected oriented surface $S$ of infinite genus is homeomorphic to a surface obtained as follows. Let $e\subset {\mathbb R}$ be a compact nowhere dense subset partitioned as $e_0\sqcup e_1$ where $e_1$ is closed and nonempty. Take any sequence of closed disks $D_j$ in ${\mathbb C} - e$ such that the accumulation set of this sequence is exactly $e_1$. Remove interiors of the disks $D_j$ and identify the boundary circles (with orientation induced from the complex plane) of the resulting surface pairwise by orientation-reversing homeomorphisms. The result is homeomorphic to your surface $S$.

In the finite genus case, the model is the essentially same, only $e_1=\emptyset$ and the collection of disks $D_j$ is finite, of even cardinality.

Given this, we can assume that our disks are chosen to have boundary circles meeting the real axis orthogonally and the circle-pairing maps $$ f_j: \partial D_j \to \partial D_{j'} $$ to be equivariant with respect to the complex conjugation: $$ f_j(\bar{z})=\overline{f_j(z)}. $$ Then the complex conjugation descends to an orientation-reversing homeomorphism of the surface $S$. qed