If $U$ commutes with $H$, can we say if U also commutes with $H^\dagger$?

Not without more hypotheses. For example specializing to the case $U = H$, the hypothesis that $U$ commute with $H$ is trivially satisfied, while the desired conclusion that $H$ commute with $H^{\dagger}$ holds only for a special class of operators (the operators known as "normal" operators, which is usually defined by this condition). Any non-normal operator is thus a counterexample.

Consider for example the operator given by the matrix $T = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ regarded as an operator on $\mathbb{C}^2$ with usual orthonormal basis $e_1 = (1,0)^T$ and $e_2 = (0,1)^T$. A short calculation shows that $TT^{\dagger} e_1 = T e_2 = e_1$ while $T^{\dagger} T e_1 = T^{\dagger} 0 = 0$ so that $TT^{\dagger}$ and $T^{\dagger} T$ are not the same.

If $H$ is assumed to be normal, then if $U$ is a bounded operator and $UH = HU$, it is also true that $UH^{\dagger} = H^{\dagger} U$. While not hard to prove, this is a nontrivial result, sometimes referred to as Fuglede's theorem. See https://en.wikipedia.org/wiki/Fuglede%27s_theorem