This is true if $M=mI_n$, and false otherwise.

The function $f$ is a gradient if its differential is symmetric, that is if
$$\langle \nabla_Yf(X),Z\rangle=\langle \nabla_Zf(X),X\rangle$$
for every $X\in S_n^{++}$ and $Y,Z\in S_n$. This amounts to
$${\rm Tr}(MX^{-1}YX^{-1}MZ)={\rm Tr}(MX^{-1}ZX^{-1}MY).$$
With teh cyclic property of the trace, and the fact that $\langle A,B\rangle={\rm Tr}(AB)$ is a scalar product, this amounts to
$$X^{-1}MZMX^{-1}\equiv MX^{-1}ZX^{-1}M.$$
This rewrites as $B^TZB=Z$ for every $Z$, where $B=MX^{-1}M^{-1}X$. This tells us that $B=I_n$. This being true for every $X\in S_n^{++}$, we obtain the necessary and sufficient condition that $M$ commuttes with every $X^{-1}$. In other words, $M=mI_n$ for some $m>0$.

What is $\phi$ when $M=I_n$ ? Nothing but $\log\det X$. As a matter of fact
$$\nabla g(\det X)=g'(\det X)\nabla\det X=g'(\det X)\hat X=g'(\det X)(\det X)X^{-1},$$
thus it suffices to have $g'(s)=1/s$.