Suppose for all value of $r$ expression for Effective Potential Energy $U_{eff}$ is zero, does that mean $F(r)$ is zero?

In general, the answer is no.

Note $U_{eff}(r)=frac{L^2}{2mu r^2}+U(r)=0$

Take derivative of both sides yield:

$frac{-L^2}{mu r^3}+nabla U(r)=0$

This gives

$F(vec{r})=-nabla U(r)=frac{-L^2}{mu r^3}$

Thus, in general only if angular momentum is zero, force of all value $vec{r}$ is zero.