The resulting formation contains 125 + 150 + 12 + 54 = 341 cubes341 cubes.
Perhaps this is not optimal for a sphere of radius 5, but the fact that each vertex of each of the nine cuboids (72 points in all) is within $(10-\sqrt{99})/2\leq0.026$ of the surface of the sphere suggests that it may be hard to beat.
Update 16 May 2017
The above solution turns out not to be optimal. Note that it arranges the cubes, along each of three orthogonal axes which will be called X, Y and Z, into nine "slices", each of one cube thickness. The configuration of the second and eighth slices along each axis (six slices in all) is as below.
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/ixc6H.jpg)
Two extra cubes can be added in each slice by sliding two rows of cubes a distance of half a cube length, producing the configuration below.
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/FbtFA.jpg)
To make this possible on each of the six faces, care is needed to avoid a change in one slice blocking a change in a face at right angles to it. One way to achieve this is:
- For the two slices in the X-Y plane, slide rows parallel to the X axis.
- For the two slices in the Y-Z plane, slide rows parallel to the Y axis.
- For the two slices in the X-Z plane, slide rows parallel to the Z axis.
In terms of overlapping cuboids, this gives three 7 x 6 x 3 cuboids, centred at the centre of the sphere, with long diagonals of length $\sqrt{94}<10$.
Altogether, this adds 6 x 2 = 12 cubes.
The first and ninth slices on each axis consist of a 3 x 3 block of cubes. In a similar way, a cube can be added to each block by sliding a central row of three cubes half a cube length. The resulting extra cuboids have dimensions 9 x 4 x 1, with long diagonal $\sqrt{98}<10$. This adds a further 6 cubes.
The formation resulting from these changes has 341 + 12 + 6 = 359 cubes.