If you take the divergence of the electric field outside the sphere, i.e. $r \gt R$, you will get get everywhere $$div \vec E=0$$ because the volume charge density $\rho=0$. You can also see this by expressing the divergence of $\vec E$ in cartesian coordinates $$\vec E=const \frac {\vec r}{|\vec r|^3}$$ $$\frac {\partial E_x}{\partial x}+\frac {\partial E_y}{\partial y}+\frac {\partial E_z}{\partial z}=0$$ for $r \gt R$.

If you take the divergence of the electric field outside the sphere, i.e. $r \gt R$, you will get get everywhere $$div \vec E=0$$ because the volume charge density $\rho=0$. You can also see this by expressing the divergence of $\vec E$ in cartesian coordinates $$\vec E=const \frac {\vec r}{|\vec r|^3}$$ $$\frac {\partial E_x}{\partial x}+\frac {\partial E_y}{\partial y}+\frac {\partial E_z}{\partial z}=0$$ for $r \gt R$.

PS: As @Sean E. Lake has rightly pointed out, for this Coulomb field its even easier to see this using the divergence in polar coordinates $$div \vec E =\frac {1}{r^2}\frac{\partial (r^2 E_r)}{\partial r}$$