7. Given $S$ is a parallelepiped. Proposition $p: S$ is a cube; Proposition $q: S$'s any body diagonal is perpendicular to any face diagonal that does not intersect it. Then proposition $p$ is a ( ) condition for proposition $q$.
(A) Sufficient but not necessary
(B) Necessary but not sufficient
(C) Sufficient and necessary
(D) Neither sufficient nor necessary
7. C.

Sufficiency is obvious.
Necessity.
As shown in Figure 3, the top and bottom faces are parallelograms, and each side face is a rectangle. $\square$

Draw the projections of the body diagonals $A_{1} C$ and $B_{1} D$ on the plane $A A_{1} B_{1} B$.

Since the plane $A A_{1} B_{1} B \perp$ plane $A B C D$, the feet of the perpendiculars from points $C$ and $D$ to the plane, $C^{\prime}$ and $D^{\prime}$, must lie on the line $A B$. The projections of $A_{1} C$ and $B_{1} D$ on the plane $A A_{1} B_{1} B$ are $A_{1} C^{\prime}$ and $B_{1} D^{\prime}$, respectively.
The figure on the plane $A A_{1} B_{1} B$ is shown in Figure 4.
Since $C^{\prime} D^{\prime} / / A_{1} B_{1}, C^{\prime} D^{\prime}=C D=C_{1} D_{1}=A_{1} B_{1}$, it follows that quadrilateral $A_{1} B_{1} C^{\prime} D^{\prime}$ is a parallelogram, and the intersection of its diagonals is point $Q$.

Also, $A_{1} C \perp A B_{1}, B_{1} D \perp A_{1} B$. By the theorem of three perpendiculars, we have
$A_{1} C^{\prime} \perp A B_{1}, B_{1} D^{\prime} \perp A_{1} B$.
Thus, $\angle A_{1} Q B_{1} + \angle A_{1} P B_{1} = \pi$.
Since the distance from point $Q$ to $A_{1} B_{1}$ is equal to the distance from point $P$ to $A_{1} B_{1}$, which is $\frac{1}{2} A A_{1}$, and $Q$ is the intersection of the altitudes from the two equal sides of $\triangle A_{1} P B_{1}$, $\triangle A_{1} Q B_{1}$ is also an isosceles triangle. Therefore, point $Q$ coincides with point $P$, and
$\angle A_{1} P B_{1} = \angle A_{1} Q B_{1} = \frac{\pi}{2}$.
Hence, the side face $A A_{1} B_{1} B$ is a square.
Similarly, the other side faces are also squares.
Since points $D^{\prime}$ and $C^{\prime}$ coincide with points $A$ and $B$, respectively, it follows that the base quadrilateral is a square.
In summary, the hexahedron $S$ is a cube.