Not every real algebraic surface can be endowned a structure of a complex algebraic curve. The only obstruction I know is orientability.
Are there any others?
Not every real algebraic surface can be endowned a structure of a complex algebraic curve. The only obstruction I know is orientability. Are there any others? 


If you're not requiring any compatibility criterion between the real and complex structure, then the only obstruction is in fact orientability. Every smooth projective real algebraic surface is a smooth compact real 2manifold (without boundary). If it's orientable, it must then be a surface of genus $g$ for some $g$. But every surface of genus $g$ admits a complex structure, and every Riemann surface is algebraic. I don't study real algebraic geometry much, but I'm not aware of a good compatibility condition to impose on your complex structure. If you've got something in mind, let me know. 


As observed in some of the previous comments, every closed (=compact, without boundary) orientable real 2manifold admits a complex structure. So in the smooth case orientability is essentially the only obstruction. If one also considers the case of singular real algebraic surfaces, the situation is more involved and I don't know whether satisfactory results are known. Anyway, one obvious obstruction is the presence of nonisolated singularities, since every complex curve has only a finite number of singular points. For instance, take $X:=S^1 \times C$, where $C \subset \mathbb{RP}^2$ is the nodal real cubic of equation $y^2z=x^3+x^2z$. The singular locus of $X$ is isomorphic to $S^1$, so $X$ surely cannot be endowed with the structure of a complex algebraic curve. 


If it is orientable, you have a complex structure and the field of meromorphic functions. Putting my ears into the firing line, I suggest that something should go wrong with the transcendence degree of the field of meromorphic functions. If it is 1, you can consider DVRs that will give you a compact algebraic curve, and I see no reason for the original curve not to be a subset. If it is more than 1 the surface cannot be algebraic... 

