Assuming $A$ itself is time-independent. If $[A, U_t]=0$ for all $t$, then it can be proven that $[A, H]=0$: We have
$H=i \partial_t U_t\cdot U_t^\dagger$
Because $[A, U_t]=0$, it follows that $[A, \partial_t U_t]=0$ since we assume $\partial_t A=0$.
Now $[A, H]=[A, H_0]+w(t)[A, H_c]=0$. Again we take derivative with respect to $t$, we get $w'(t)[A, H_c]=0$. So we conclude that as long as $w'(t)\neq 0$ for some $t$, $[A, H_c]=[A, H_0]=0$.