First, R and r are proportional, by dimensional analysis, so set r to 1. Set g to 1 by choosing the unit of time to be $\sqrt{r/g}$.
The ice-cream height at any time is $2 - {t^2\over 2}$, where time starts when everything is at the top of the loop, the first instant of falling. So the ice-cream gets to the bottom at t=2.
At angle $\theta$ from the top, the velocity of the car is, by conservation of energy, $\sqrt{2(R + 1 - \cos(\theta))}$. The time taken to drop is the integral of the distance over the velocity along the circle, or
$$ {1\over \sqrt{2}} \int_0^\pi {d\theta \over \sqrt{R - 1 - \cos(\theta)}} $$
And setting this time equal to 2 gives a transcendental equation for H=R-1 whose solution is the answer (and extend the domain of integration to the full circle).
$$ \int_0^{2\pi} {d\theta \over \sqrt{H - \cos(\theta)}} = 4\sqrt{2} $$
expand the denominator in powers of $\cos(\theta)$ using the power series for square-root below (a simple Taylor series at x=0):
$$ (1-x)^{-{1\over 2}} = \sum_{N=0}^{\infty} {(2N)!!\over 2^N} {x^N \over N!}$$
Where the (2N)!! is a weird notation product of odd numbers less than 2N, which is given by:
$$ 2N!! = 1\cdot 3\cdot 5 ... \cdot (2N-1) = {(2N)!\over 2^N N!}$$
and use this useful identity for integer N's:
$$ \int_0^{2\pi} (\cos(\theta))^{2N} d\theta = 2\pi {2N \choose N} {1\over 2^{2N}}$$
$$ \int_0^{2\pi} (\cos(\theta))^{2N+1} d\theta = 0 $$
Which is derived by writing cosine as a sum of complex exponentials and multiplying out the product, and noting that only the middle term for even powers survives. This gives a good expansion for the function of R on the left.
The function is monotonic decreasing from infinity at $H=1$ to 0 at $H=\infty$, so there is a unique solution. The resulting expansion is
$$ {1\over \sqrt{H}} \sum_{N=0}^{\infty} { (4N)!!\over (2^N N!)^2 } {1\over H^{2N}} = {2\sqrt{2}\over \pi} $$
And the solution is (by hand) very close to H=1.50, meaning that R is close to 2.50. You can invert the power series, or just solve this in a second with a computer.
If I'm understanding your problem correctly, then the normal force is the centripetal force.
In other words, the normal force from the rail causes the centripetal acceleration towards the center of the circle. There are, as I understand it, no other forces acting in the normal direction. Remember that you are only supposed to consider forces in the normal direction:
The gravitational force is perpendicular to the normal force at this position and so has no effect in the normal direction.
Best Answer
As it turns out, you actually can use that same formula $ v_{min} = \sqrt{gR} $. However $R$ is the radius of curvature at the top of the loop, which is simply equal to the radius in the case of a circle. See here for more information on finding the curvature of an ellipse.
In general, the curvature of a plane curve given by $ (x(t),y(t)) $ can be found using the formula $ \kappa = \frac{|\dot x \ddot y - \dot y \ddot x|}{(\dot x^2 + \dot y^2)^\frac 3 2} $ (or one of many other formulae) where $ R = \frac 1 \kappa $