Maybe there's a way to do it like that, it certainly gets the right answer, but that would not normally be considered the way to go. The 3D harmonic oscillator has only 3 degrees of motional freedom, and they are all vibrational. I think you are thinking of diatomic molecules, and their vibrational degree of freedom is often not excited, so they only "count" as 5 degrees of freedom in most applications. But you get the right answer of 3kT-- the reason is that each of the 3 degrees of freedom counts twice, once in momentum space and once in position space. That's because a different momentum in each direction counts as a different state, and a different position in each direction also counts because there is potential energy associated with those displacements-- free particles have a different way of counting positional states, it depends on the volume allowed per particle.
Maybe the best way to see this is the full result. The average energy is a probability weighted sum over all the possible states of the oscillator, where the relative probability of each state is the Boltzmann factor, e-(px2+py2+pz2+x2+y2+z2)/kT. (where for simplicity I've scaled p2 to 2m and x2 to 2/k, rescaling by a constant is no big thing). Counting states in phase space is like integrating over phase-space volume, so that's over dpxdpydpzdxdydz, and the weighted average of the energy requires that we multiply the integrand in the numerator by the energy, px2+py2+pz2+x2+y2+z2. The denominator is the normalization of the weighted sum, so is exactly like the integral in the numerator but without that last energy factor. Each of these integrals is completely separable, so just do a change of variables in each one that sends either something that looks like p2/kT or like x2/kT into a dummy variable like z2. The energy factor that is only in the numerator then ends up looking like 6kTz2, and the action of the integration acts only on the z variables. 5 of the 6 integrals always cancel top and bottom, and the one that doesn't cancel gives 1/2. So you end up with 6kT/2 = 3kT. Thus it is often said that every degree of freedom that shows up as a quadratic in the energy contributes kT/2 to the average energy.
Note that neither k nor m matters, we scaled them out and they never came back. That's why your way might be equivalent, you are in effect doing it in spherical rather than Cartesian coordinates, and scaling out the m and k makes it ambiguous as to which are kinetic energies and which are potential. Your 3 degrees of translational freedom map into my 3 degrees of momentum, and your 2 degrees of rotational energy map into 2 of my vibrational degrees of freedom. I think the correspondence might just be coincidental, however.
