Well, you have asked a clever and different question out here. I would try solving your doubt.
In all the cases the intial energy is same i.e. mgh.
Now, for all the three bodies, the translational motion is exactly the same. Only the rotational motion differs. Hence they will have exactly the same final translational K.E. and frictional force being same in all the cases i.e. µMg cos theta, and the distance travelled by all the bodies is same. Thus, Work done against friction is also same in all the cases.
Equating Ktrans. + Krot. + WD against friction = Mgh in all the cases,
Since, Ktrans + WD against friction is same in all the cases, therefore, Krot. will be automatically same for all the three bodies. Since, all the parameters are same for all the bodies, hence, they will reach at the same time.
I don't whether my answer is correct or not, if anyone gets a different answer please reply.