Put $n=m+r$, and then we can write $R(m,r)$ more conveniently as 
$$ 
R(m,r) = \frac{(2n)!}{m! (n+r)!} \frac{m! r!}{n!} = \frac{\binom{2n}{n} }{\binom{n+r}{r}}. 
$$ 
So the question essentially becomes one about which numbers $n+k$ for $k=1$, $\ldots$, $r$ divide the middle binomial coefficient $\binom{2n}{n}$.   Obviously when $k=1$, $n+1$ always divides the middle binomial coefficient, but what about other values of $k$?  This is treated in a lovely [Monthly article of Pomerance][1].  

Pomerance shows that for any $k \ge 2$ there are infinitely many integers with $n+k$ not dividing $\binom{2n}{n}$, but the set of integers $n$ for which $n+k$ does divide $\binom{2n}{n}$ has density $1$.  So for any fixed $r$, for a density $1$ set of values of $n$ one has that $(n+1)$, $\ldots$, $(n+k)$ all divide $\binom{2n}{n}$, which means that their lcm must divide $\binom{2n}{n}$.  But one can check without too much difficulty that the lcm of $n+1$, $\ldots$, $n+k$ is a multiple of $\binom{n+k}{k}$, and so for fixed $r$ one deduces that $R(m,r)$ is an integer for a set of values $m$ with density $1$.   (Actually, Pomerance mentions explicitly in (5) of his paper that $(n+1)(n+2)\cdots (n+k)$ divides $\binom{2n}{n}$ for a set of full density.)

 I haven't quite shown that $R(m,r)$ is not an integer infinitely often for $r\ge 2$, but I think this can be deduced from Pomerance's paper (by modifying his Theorem 1). 

[1]: https://www.math.dartmouth.edu/~carlp/amm2015.pdf