Cone flux!

According to Gauss's law the total flux on the surface (including the base) is

$\Phi=\frac{q}{\epsilon_0}$

We divide this to two parts: $\Phi= \Phi_C+\Phi_B$ , where $\Phi_C$ is the flux though the curved surface and $\Phi_B$ is the flux through the base. Clearly, it is easier to calculate $\Phi_B$ , and we can use $\Phi_C= \Phi-\Phi_B$ to answer the original question.

Simple geometry tells us that the line connecting the charge to the perimeter of the base makes an angle of $\theta_0=\cos^{-1}\frac {h}{\sqrt{h^2+16R^2}}$ with the vertical. The solid angle covered by the base is:

$\Omega= \int_0^{\theta_0} \int_0^{2\pi} \sin \theta d \theta d \phi = 2\pi(1-\cos \theta_0)$

The full flux $\Phi$ is evenly distributed over a solid angle of $4\pi$ . The solid angle belonging to the curved surface is $4\pi-2\pi(1-\cos \theta_0)= 2\pi(1+\cos \theta_0)$ . Therefore the flux is

$\Phi_C=\frac{q}{\epsilon_0}\frac{ 2\pi(1+\cos \theta_0)}{4\pi}= \frac{q}{2} \left(1+\frac {h}{\sqrt{h^2+16R^2}}\right)$