"Yes" for the case $m=n=3$.
I will assume that center of sphere lies in the first polyhedron, say $Q$, but it is straigthforward to remove this assumption.

Assume contrary. 
By rescaling the second polyhedron, say $P$, we can assume that the radius of the sphere stays the same as for the original polyhedron $Q$, but all edges get shorter.

Consider the central projection of the edges of $Q$ to the sphere.
Note that they cut the sphere into cyclic polygons.
Choose an edge $e$ of $Q$, consider two of its adjacent spherical cyclic polygons $F$ and $G$. 
The vertices of $F$ and $G$ lie on two arcs with endpoints of $e$.
Since $Q$ is convex the arcs bound a nonconvex region from the sphere with $F$ and $G$ inside.
The last statement (with some work) implies the following: if one decreases the edge $e$ while keeping $F$ and $G$ cyclic, then the total area of $F$ and $G$ decreases.

Now, for each edge of $Q$ draw a line segment between the corresponding vertices of $P$.
Project the obtained line segments to the sphere.
Since the center of sphere is inside of $P$ the spherical polygons that correspond to facets of $Q$ will cover the whole sphere.
Note that cyclic polygon maximize the area among all polygons with the same sides.
From above it follows that the total sum of their areas get smaller, but they cover the same sphere --- a contradiction.