Solution:
Imagine drawing the sphere and the cube. Take a cross section, with a plane parallel to two of the cube's faces, passing through the sphere's center. In this cross section, the sphere looks like a circle, and the cube looks like a square (of side length $1$) inscribed in that circle. We can now calculate that the sphere has diameter
$$d := \sqrt{2}$$
and surface area
$$S := \pi d^{2} = 2 \pi,$$
and that the sphere protrudes a distance of
$$x := \frac{\sqrt{2} - 1}{2}$$
out from any given face of the cube.

It is known that the surface area chopped off from a sphere by any plane is proportional to the perpendicular distance thus chopped off. Thus, each face of the cube chops off a fraction
$$\frac{x}{d}$$
of the sphere's surface. The surface area of the pebble contributed by the sphere is thus
$$S \cdot \left(1 - 6 \cdot \frac{x}{d}\right),$$
whereas the cube contributes $6$ circles of radius $\frac{1}{2}$, with total area
$$6 \cdot \pi \left(\frac{1}{2}\right)^{2} = \frac{3}{2} \pi.$$
The pebble's surface area is therefore
$$S \cdot \left(1 - 6 \cdot \frac{x}{d}\right) + \frac{3}{2} \pi = 2 \pi \cdot \left(1 - 6 \cdot \frac{\sqrt{2} - 1}{2 \sqrt{2}}\right) + \frac{3}{2} \pi = \frac{6 \sqrt{2} - 5}{2} \pi.$$