There is something peculiar about this question.
Let us calculate the torque equation about the center of mass of the rolling sphere:
$$\tau = f_sr = I\frac{d\omega}{dt} = mr^2\frac{d\omega}{dt} = mr\frac{dv}{dt} $$ since $r\omega = v$ according to rolling condition.
This shows us that the force of friction $f_s$ will never be zero as long as the rolling sphere is perfectly rolling on the surface of the fixed sphere.
In other words, as long as you use the rolling condition in your solutions you cannot equate $f_s$ to zero!
Therefore, this seems to me a very artificial and unrealistic problem.
This is what one would expect in actual conditions: the normal reaction offered by the fixed sphere on the rolling sphere will keep reducing as the sphere moves lower and lower and will reach zero when the $cos\theta$ component of gravity just equals the centripetal force $\frac{mv^2}{R+r}$. It is at this point that the two spheres lose contact. While the force of friction which has its maximum value $\mu N$ would have long before loss of contact decreased to such an extent that it would no longer be able to maintain pure rolling.
Of course, you can insist that you still do the question as it is given. This is how you can do it:
The torque equation about the point O is (at all times while the spheres are in contact):
$$m(R + (1+\beta)r)\frac{dv}{dt} = mg(R+r)sin\theta - fR$$
Applying Newton's second law:
$$m\frac{dv}{dt} = mgsin\theta - f$$
Now let us eliminate the $f$ term:
$$m(R + (1+\beta)r)\frac{dv}{dt} = mgrsin\theta + Rm\frac{dv}{dt}$$
$$m(1+\beta)r\frac{dv}{dt} = mgrsin\theta $$
which finally gives:
$$(1+\beta)\frac{dv}{dt} = gsin\theta $$
Now,
$$\frac{dv}{dt} = \frac{dv}{d\theta}.\frac{d\theta}{dt} = \frac{dv}{d\theta}.\frac{v}{R+r}$$
which gives you the differential equation:
$$\frac{(1+\beta)}{R+r}.\frac{vdv}{d\theta} = gsin\theta$$
Integrate to get,
$$\frac{(1+\beta)}{R+r}.\frac{v^2}{2} = -gcos\theta + c$$
At $\theta=0$, $v=0$ and at the required $\theta$, $mgcos\theta = \frac{mv^2}
{R+r}$
Solve further, and you get the result that you may have got from work-energy considerations that:
$$cos\theta = \frac{2}{3+\beta}$$
