No, this is not correct
Certainly the ring will move tangent to what would have been its trajectory at the point of release had it not been released. But as the ring moves outward it will not be maintaining a constant distance from the fixed point of the rod. Therefore, its radial position $r$ will be changing, i.e. it has a non-zero radial velocity still.
You can determine the equation of motion by looking at Newton's second law for planar motion in polar coordinates:
$$\mathbf F=m\mathbf a=m(\ddot r-r\dot\theta^2)\hat r+m(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$
Since there is no force acting on the ring after it leaves the rod, it must be that each component of the acceleration is $0$:
These equations, along with the initial conditions you describe, determine the motion of the ring after it leaves the rod.
You have already convinced yourself that after the ring leaves the rod, $\dot\theta\neq0$. And we know that $r\neq0$. Therefore $\ddot r\neq0$, and so it must be that there is a non-zero $\dot r$.
Therefore, to answer your title question
Will there be any radial velocity in absence of centripetal force and angular acceleration?
Yes, there will be. It is determined by the motion of the ring once it reaches the end of the rod.