Formally it’s easy to bring counterexample: suppose $f(1)=1$, $g(1)=0$ and then, for other values of argument, $f, g$ are any pair of non-zero functions with property $f(n)in O(g(n))$. Now taking $h(n)=1,forall n in mathbb{N}$ makes impossible $f(h(n)) in O(g(h(n)))$.

On other hand, for example, if $h$ is strictly increasing function, then your claim will be true, because we obtain property for subsequence from sequence.