Brian's solution remains valid if we make a slightly more realistic assumption by ignoring dissipative friction but taking friction to be strong enough to cause the ball to roll rather than slide. In this case the ball's energy is given by
$ \begin{align} E &=\frac12mv^2+\frac12I\omega^2+mgz \\ &=\frac12(1+\alpha)mv^2+mgz\;, \end{align} $
where the factor $\alpha=I/(mR^2)$, with $I$ the ball's moment of inertia and $R$ its radius, depends on the mass distribution and is $\frac25$ for a solid ball of constant density.
Thus the rotational energy effectively increases the ball's inertial mass, or equivalently reduces the gravitational acceleration. The solution is still a cycloid; only the time required to reach the ground is now
$ T=\pi\sqrt\frac{r(1+\alpha)}g $
(where $r$ is the radius of the cycloid), i.e. the rolling ball takes slightly longer than the sliding ball, by a factor of $\sqrt{7/5}\approx1.18$ in the case of a solid ball of constant density.
Note that this treatment ignores the fact that for a ball with non-zero radius $R$ the ball's centre of mass moves on a different curve than its surface; this approximation is only valid for $r\gg R$.