Not the simple substitution. The problem changes completely. Now you have all mass in the shell, so is, almost each piece of the sphere farther than before from the axis and it mounts that the farther from the shell before, the more contributing to the moment of inertia now. But, it's true now as before that the tensor is diagonal, so it's only needed an integral about any axis passing by the center. Consider the sphere centered in the origin and calculate the moment of inertia around the $z$ axis. We can do it better in spherical coordinates.
The distance from any point of the shell to the $z$ axis is $r\sin\phi$
and the (constant) density of mass $\rho$
$$I_z=\int_{R}^{R+a}\int_0^\pi\int_0^{2\pi}r^2\sin^2\phi r^2\sin\phi\,\rho\,\mathrm d\theta\mathrm d\phi\mathrm dr=$$
$$=\rho\int_{R}^{R+a}r^4\int_0^\pi\sin^3\phi\int_0^{2\pi}\mathrm d\theta\mathrm d\phi\mathrm dr=$$
$$=2\pi\rho\int_{R}^{R+a}r^4\int_0^\pi\sin^3\phi\,\mathrm d\phi\mathrm dr=$$
$$=2\pi\rho\int_{R}^{R+a}r^4\left[(1/12)(\cos3\phi-9\cos\phi\right]_0^\pi\mathrm dr=$$
$$=\frac 83\pi\rho\int_{R}^{R+a}r^4\mathrm dr$$
Now, we can apply the condition $a\ll R$. Define $K(x)=\int_{R}^{R+x}Rr^4\mathrm dr$
$$K(x)=\int_{R}^Rr^4\mathrm dr+R^4x+O(x^2)=R^4x+O(x^2)$$
Then, for this shell:
$$I_z\approx\frac 83\pi\rho R^4a$$
Now, the shell has mass $M$ and in a symilar way as for the moment of inertia, we have the volume of the shell is in good approximation $V=4\pi R^2a$ and the mass is $M=4\pi\rho R^2a$. Finally, we can write:
$$I_z=\frac 23MR^2$$
Very different from the simple substitution.
If you are including the motion of the CM is because you are considering the motion of each part of the rod as composition of two motions: the motion of each part around the CM and the motion of the CM with respect to our frame (in wich $O$ is at rest). In this case, the moment of inertia has to be around the CM: $I = \dfrac 1{12} ML^2$ leading to
$$T=\frac 18 ML^2 \dot \theta ^2 + \frac 1{24} ML^2 \dot \theta ^2=\frac 16 ML^2 \dot \theta ^2$$
We have two contributions: kinetic energy of rotation around the CM and kinetic energy of translation of the CM.
If we choose to consider the parts of the rod as simply moving around $O$ we have to use the moment of inertia around $ O$ because there are no composition of motions for any part and there is only energy of rotation: $T=\dfrac 16 ML^2 \dot \theta ^2$
Best Answer
General formula for KE of rotating body is
$$K = \frac{1}{2} \ I \omega ^2+\frac{1}{2}mv_{\text{com}}^2$$
where
$\omega=\dfrac{v}{r}$ for rolling without slipping, and
$v_{\text{com}} $ is velocity of centre of mass (here $v_{\text{com}}=v$)
Thus for spherical shell,
$$K = \frac{1}{2} \cdot \dfrac{2mr^2}{3}\cdot \dfrac{v^2}{r^2}+\frac{1}{2}mv^2$$
$$\therefore K = \dfrac{5}{6}mv^2$$
Regarding your intuition, since both the rotating bodies have same velocity and mass, then moment of inertia determines which body will have greater $K$.
Moment of inertia is rotational analogue of translational inertia, i.e. mass. When two bodies have same velocity, then the body with higher mass will have greater $K$.