If you change $t\mapsto -t$, equation does not change, and $u|_{t=0}$ does not change, but $u_t|_{t=0}$ acquires sign "$-$". However, since $u|_{t=0}=0$, if you replace in addition $u\mapsto -u$, then nothing changes. Thus $u(x,t)=-u(x,-t)$.

If on the other hand, $h(x)=0$ then $u$ would be even with respect to $t$