Trace of non-commutable matrices

Your conjecture is a special case of the following result which essentially follows from the Lieb-Thirring inequality.

Let $A$ and $B$ be Hermitian matrices. Then, for every positive integer $p$ we have \begin{equation*} |\text{tr}(AB)^{2p}| \le \text{tr}A^{2p}B^{2p} \end{equation*}

Not to take anything away from Suvrit's answer, but this is actually much simpler. First, we can assume $M_1$ is diagonal. Call it $diag(x_1, \dotsc, x_i).$ Then the difference between the LHS and the RHS is

$$\sum_{i> j} a_{ij}^2 (x_i - x_j)^2,$$ where the $a_{ij}$ are the entries of $M_2.$