Solution:
One can show that the hexagon must have as its vertices the midpoints of six edges of the cube, as illustrated; for example, this readily follows from the fact that opposite sides of the hexagons and the medians between them are parallel. We then conclude that the side of the hexagon is $\sqrt{2} / 2$ (since it cuts off an isosceles triangle of leg $1 / 2$ from each face), so the area is $\left(\frac{3}{2}\right)\left(\frac{\sqrt{2}}{2}\right)^{2}\left(\sqrt{3}\right) = \frac{3 \sqrt{3}}{4}$. Also, the plane passes through the center of the sphere by symmetry, so it cuts out a cross section of radius $1 / 2$, whose area (which is contained entirely inside the hexagon) is then $\pi / 4$. The sought area is thus $\frac{3 \sqrt{3} - \pi}{4}$.