No. It is known, for instance, that $B\ge A\ge0$ does not imply $B^2\ge A^2$. Now pick such a pair of real matrices $A$ and $B$. Then $u^T B^2u<u^T A^2u$ for some nonzero real vector $u$. Let $M=uu^T$. Then
$$
\|BMB\|_2= u^TB^2u<u^TA^2u=\|AMA\|_2.
$$