If $X$ is closed then $JX$ is closed Let $(S, g, J)$ be a closed Riemann surface with a Riemannian metric $g$ compatible with the complex structure $J$. Suppose that a smooth vector field $X$ on $S$ is closed, i.e., the $1$-form $\omega = g(X, \cdot)$ is closed. Is it true that the $1$-form $J \omega := g(JX, \cdot)$ is also closed?
I suspect so, but I don’t know a proof.
Added: this is not true if $S$ is a sphere (as pointed by @Aloizio Macedo), because $H^1(S)=0$ in this case. If we impose that $S$ is not a sphere, does my statement hold?
 A: No. First let me change $\omega$ to $\eta$ so I don't get confused. 
Let $S$ be the sphere with $J$ being rotation on the tangent plane around the normal vector by $\pi/2$. If $H$ is the height function and $X=\nabla H$, then $\eta=dH$, hence closed.
Since $H^1(S)=0$, all closed forms will be exact. So, if $g(JX,\cdot)$ were closed, it would be exact and then $g(JX,\cdot)=dF$ for some $F$. Since $JX$ has non-trivial closed orbits, this is an absurd. Indeed, letting $\gamma$ be such a closed orbit,
$$ \int_{\gamma}dF=F(\gamma(T))-F(\gamma(0))=0,$$
but 
$$\int_{\gamma}dF=\int_0^TdF(\dot{\gamma})=\int_0^Tg(JX,JX)>0.$$
EDIT: OP asks if this is true in case of $S$ not being the sphere. No, $H^1(S)$ being zero was a way to simplify things. Take the torus and the height function $H$ again, with $J$ being again rotation around the normal vector. As before, $X=\nabla H$ makes $\eta=dH$, hence closed.
$g(JX,\cdot)$ is equal to $f\alpha$, where $\alpha$ is one of the generators of the torus cohomology (hence, closed) and $f$ is nonconstant but constant on horizontal circles. (Note that $f$ decreases to $0$ near the top and bottom of the torus.) Hence, $d(f\alpha)=(df)\alpha+f(d\alpha)=(df)\alpha$. You can compute this to see that it is non-zero. For instance, 
$$((df)\alpha)(\partial_{\theta_1}, \partial_{\theta_2})= df(\partial_{\theta_1})\alpha(\partial_{\theta_2})-df(\partial_{\theta_2})\alpha(\partial_{\theta_1})=-df(\partial_{\theta_2})\alpha(\partial_{\theta_1})=-df(\partial_{\theta_2}) \neq 0$$
if you are over some point which is not on the equator or the top/bottom, for example.
A: Another way to see this is the following: Since $J$ is compatible with $g$, we have 
$$ g(X, JX) = 0$$
for all $X$. In particular, $J\omega = *\omega$, where $*$ is the Hodge star operator with respect to $g$. Thus given the closed form $\omega = g(X, \cdot)$, $J\omega$ is closed if and only if $*\omega$ is closed. This is true if and only if $\omega$ is a harmonic one form. 
