The moment of inertia is defined relative to an axis, not a point. Therefore the distance you need is from a mass element in the sphere to the $z$-axis, not to the origin. This is why the disk method is used -- it takes advantage of the symmetry. Your expression uses the distance $r$ to the origin, whereas the distance to the $z$ axis in spherical coordinates is $r\sin \theta$ (where $\theta$ is the angle between the $z$ axis and the point of interest).
The correct expression for a spherical object with arbitrarily varying density in spherical coordinates is
\begin{eqnarray}
I &=& \int_0^R {\rm d} r r^2 \int_0^{\pi} {\rm d\theta} \sin \theta \int_0^{2\pi}{\rm d}\phi \rho(r,\theta,\phi)\Delta(r,\theta,\phi)^2,
\end{eqnarray}
where $\rho(r,\theta,\phi)$ is the density and $\Delta(r,\theta,\phi)=r\sin\theta$ is the distance from a mass element at $r,\theta,\phi$ to the $z$ axis.
For completeness we can work this out for a constant density sphere, $\rho(r,\theta,\phi)=3 M/(4\pi R^3)$ (note that the integrand doesn't depend on $\phi$, so the $\phi$ integral is just $2\pi$)
\begin{eqnarray}
I &=& 2\pi \frac{3M}{4\pi R^3} \int_0^R {\rm d} r r^2 \int_0^{\pi} {\rm d\theta} \sin \theta\ \left(r \sin\theta\right)^2 \\
&=& \frac{3 M}{2 R^3} \int_0^R {\rm d} r r^4 \int_0^\pi {\rm d}\theta \sin^3 \theta \\
&=& \frac{3 M}{2 R^3} \times \frac{R^5}{5} \times \frac{4}{3} \\
&=& \frac{2}{5} MR^2
\end{eqnarray}
Best Answer
As you can see, $\rho$ only depends on a single variable: $r$. Thus, it should be intuitive that one can do this problem by integrating only over the variable $r$.
To see what you are supposed to do, consider what happens if you fix $r$: You obtain a spherical shell (as was pointed out in the comments). The moment of inertia of a spherical shell is quite easy to calculate. Once you've done that, use the idea that a ball can be viewed as the infinite sum of thin spherical shells of varying radius $r$ (this is where the integration comes in). You should be able to obtain an answer without performing a double integral.