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.)
Added: Here is an example that is not even a Dold cofibration. Let $Y$ be the subspace of $\mathbb{R}^2$ that is the union of the line segment connecting $(0, 1)$ to $(0, 0)$ and the line segments connecting $(0, 1)$ to $(2^{-m}, 0)$ for all $m \in \mathbb{N}$. The inclusion $\{ 0 \} \to Y$ is a closed embedding, again between strongly contractible spaces. If it was a Dold cofibration, then it would admit a strong deformation retraction (this should be somewhere in the reference I mention in the comment). However, it is a standard exercise in topology that no such retraction exists.