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 $K$.

Edit: as pointed out by Felipe below, what used to be here was sort of wrong-headed, but I can't delete this answer since it has already been accepted. Anyway you should do what he says: the divisor of all Weierstrass points counted with weights is defined by the vanishing of the Wronskian determinant of a basis of the space of $1$-forms on $X$. This divisor is then manifestly defined over $k$ and so all the Weierstrass points are defined over a finite extension.