Probably you want $g > 1$. 


The answer is yes since there are only finitely many Weierstrass points. Explicitly, if $X$ is defined over $k$ then $\mathrm{Gal}(\overline k/k)$ acts on the set of Weierstrass points of $X$, giving a homomorphism $\mathrm{Gal}(\overline k/k) \to \mathrm{S}_n$ for some $n$. The kernel is a finite index subgroup corresponding to an algebraic extension $K/k$ such that each Weierstrass point is defined over $K$.