Let's say that the sphere has radius 1 centered at the origin in $\Bbb R^3$ and the disks are moving with speed 1. And let's also say that in the picture the $z$-axis points upward, and the $y$ axis points normal to the plane of motion of the black thing. Let's also use the convention that at time zero the black disk has center at $(0,0,1)$ and the white one at $(0,1,0)$ Then You can literally parametrize the two curves which describe the center of the disks in a simple way:

$\gamma_{black}(t) = (\sin t, 0, \cos t)$

$\gamma_{white}(t) = (-\sin t, \cos t, 0)$

So if you compute distances (by subtracting and taking the norm squared), it is clear that they are not always equidistant.

So why does it seem like they are always equidistant in the picture?

Well, if you take a good look, the motion of the two disks is not along an exact sphere but more like some kind of ellipsoid.

So instead of using the spherical model $x^2+y^2+z^2=1$, try the ellipsoidal model $x^2+y^2/2+z^2/2=1$. In other words, let's consider the image of this rigid motion under the map $(x,y,z) \mapsto (x, \sqrt 2 y, \sqrt 2 z)$. So the new parametrizations will be

$\gamma_{black}(t) = (\sin t, 0, \sqrt 2 \cos t)$

$\gamma_{white}(t) = (-\sin t, \sqrt 2 \cos t, 0)$

And in this model we see that they are always equidistant! (Well those are my two cents at least. Might be total BS. Really it depends on how you want to interpret this two dimensional projection of a three dimensional motion - for all we know the planes of motion may not even be orthogonal, which would make the formulas more complicated).