No. Let $X$ be an uncountable set and consider $I^X$ with the product topology. Then the inclusion $\{ 0 \} \to I^X$ is a closed embedding between (strongly) contractible spaces which are therefore mixed cofibrant. However, this map is not a Hurewicz cofibration since it would follow that $\{ 0 \}$ is the zero set of a continuous function $I^X \to I$ which would contradict the fact that $0$ has no countable neighbourhood basis in $I^X$. In particular, this map is not a mixed cofibration. (However, it is a _Dold cofibration_ so pushouts along it are still homotopy pushouts with respect to both homotopy equivalences and weak homotopy equivalences.)