Answer
No.
Work Step by Step
When $f$ is odd, we can find that:
$f(g(-x)) = f(-g(x))= -f(g(x)) = -h(x)$
Therefore $h$ is odd.
However, when $f$ is even, we can find that:
$f(g(-x)) = f(-g(x))= f(g(x)) = h(x)$
Therefore $h$ is even.
Thus, $h$ is not always an odd function.
