No, even if $X=\mathbb R^2$.

Let $A_1$ be (the convex hull of) 4 points with coordinates $(\pm 1,\pm 1)$. Then $m(A_1)=(0,0)$, as the 4 points are on the circle $S_1$ of radius $\sqrt 2$ centered at $(0,0)$. Shift $S_1$ a small distance $\varepsilon$ in the horizontal direction, denote the resulting circle by $S_2$. For each vertex of $A_1$, mark its nearest point on $S_2$. The marked points are vertices of a rectangle $A_2$ inscribed in $S_2$. So $m(A_2)=(\varepsilon,0)$ but the Hausdorff distance between $A_1$ and $A_2$ is $\approx\varepsilon/\sqrt 2$.

No, even if $X=\mathbb R^2$.

Let $A_1$ be (the convex hull of) 4 points with coordinates $(\pm 1,\pm 1)$. Then $m(A_1)=(0,0)$, as the 4 points are on the circle $S_1$ of radius $\sqrt 2$ centered at $(0,0)$. Shift $S_1$ a small distance $\varepsilon$ in the horizontal direction, denote the resulting circle by $S_2$. For each vertex of $A_1$, mark its nearest point on $S_2$. The marked points are vertices of a convex quadrangle $A_2$ inscribed in $S_2$ and containing its center $(\varepsilon,0)$. Hence $m(A_2)=(\varepsilon,0)$ but the Hausdorff distance between $A_1$ and $A_2$ is $\approx\varepsilon/\sqrt 2$.