If $h$ and $g$ are holomorphic it seems like the answer is no. Let $f(z) = f(re^{i\theta}) = \sqrt re^{i\theta/2}$ for $\theta \in [0,2\pi)$, and let $g(z)=z^2$. Then $f$ is discontinuous on the non-negative real axis but $g(z)$ and $h(z) = z$ are holomorphic.
If $g$ and $f$ are holomorphic the answer is yes, because $h$ is holomorphic by the chain rule.
If $h$ and $f$ are holomorphic I think the answer is yes: if $h(z)=g(f(z))$ is defined in an open set $\Omega$, then $g$ is holomorphic in $f(\Omega)$. Proof:
For any $z_0\in \Omega$ we have $$\lim_{z\to z_0} \frac{h(z)-h(z_0)}{z-z_0} = \lim_{z\to z_0} \frac{g(f(z))-g(f(z_0))}{f(z)-f(z_0)}\frac{f(z)-f(z_0)}{z-z_0}$$
so that if $f'(z_0) \ne 0$
$$\lim_{z\to z_0} \frac{g(f(z))-g(f(z_0))}{f(z)-f(z_0)} = \frac{h'(z_0)}{f'(z_0)}$$
Let's say I first show that if $f'(z_0) \ne 0$, then $g$ is holomorphic at $y_0 = f(z_0)$. By the holomorphicity of $f'$ its zeros are isolated; and since a holomorphic map is open, I'd have shown that $g$ is holomorphic everywhere in $f(\Omega)$ except possibly at an isolated set of points. If I can then show that $g$ is continuous in all of $f(\Omega)$, then those points are removable singularities, so $g$ is holomorphic in all of $\Omega$.
So pick $y_0 = f(z_0) \in f(\Omega)$ such that $f'(z_0) \ne 0$. Choose $\delta$ such that $\left|\frac{g(f(z))-g(f(z_0))}{f(z)-f(z_0)}-\frac{h'(z_0)}{f'(z_0)}\right|<\epsilon$ for $z \in B_\delta(z_0)$. $f$ is holomorphic and thus an open map, so $f(B_\delta(z_0))$ is open, so we can find $\delta'$ such that $B_{\delta'}(y_0) \in f(B_\delta(z_0))$. Then substituting $y=f(z)$, we have $\left|\frac{g(y)-g(y_0))}{y-y_0}-\frac{h'(z_0)}{f'(z_0)}\right|<\epsilon$ for every $y \in B_{\delta'}(y_0)$. So $g$ is holomorphic at every such $y_0$.
Lastly, to show that $g$ is continuous, the argument is very similar. Given $y_0$, choose any $z_0$ such that $f(z_0) = y_0$. Then $\lim_{z\to z_0} \left|g(f((z))-g(f(z_0))\right| = \lim_{z\to z_0} \left|h(z)-h(z_0)\right| = 0$. Given epsilon there is a $B_\delta(z_0)$ such that $\left|g(f((z))-g(f(z_0))\right|<\epsilon$ for $z \in B_\delta(z_0)$. $f$ is open, so $f(B_\delta(z_0))$ is open, so there is a $B_{\delta'}(y_0) \in f(B_\delta(z_0))$. For every $y \in B_{\delta'}(y_0), \left|g(y)-g(y_0)\right|<\epsilon$.
Is this correct? Is there an easier way to show this?