For each natural $n$, let $R_n$ be random variable with values in $[0,1]$ such that $nR_n$ converges in distribution, as $n\to\infty$, to a random variable $X$ with a continuous cdf.

Let $T_n:=(1-R_n)^{-(n-1)/2}$. Take any real $t>1$. Then for all large enough natural $n$ $$t^{-2/(n-1)}=e^{-2\ln t/(n-1)}<1-\frac{\ln t}n$$ and hence $$P(T_n>t)=P(nR_n>n(1-t^{-2/(n-1)})) \\ \le P(nR_n>\ln t)\underset{n\to\infty}\longrightarrow P(X>\ln t) \underset{t\to\infty}\longrightarrow0.$$ So, $$\lim_{t\to\infty}\lim_{n\to\infty}P(T_n>t)=0;$$ that is, $(1-R_n)^{-(n-1)/2}=T_n=O_P(1)$, as desired.

The condition that $X$ has a continuous cdf (which actually holds in your setting) is not needed here, but with it the proof gets simpler just a little bit.

For each natural $n$, let $R_n$ be random variable with values in $[0,1]$ such that $nR_n$ converges in distribution, as $n\to\infty$, to a random variable $X$ with a continuous cdf.

Let $T_n:=(1-R_n)^{-(n-1)/2}$. Take any real $t>1$. Then for all large enough natural $n$ $$t^{-2/(n-1)}=e^{-2\ln t/(n-1)}<1-\frac{\ln t}n$$ and hence $$P(T_n>t)=P(nR_n>n(1-t^{-2/(n-1)})) \\ \le P(nR_n>\ln t)\underset{n\to\infty}\longrightarrow P(X>\ln t) \underset{t\to\infty}\longrightarrow0.$$ So, $$\lim_{t\to\infty}\lim_{n\to\infty}P(T_n>t)=0;$$ that is (cf. e.g. equivalence (i)$\iff$(v) in Lemma 1), $(1-R_n)^{-(n-1)/2}=T_n=O_P(1)$, as desired.

The condition that $X$ has a continuous cdf (which actually holds in your setting) is not needed here, but with it the proof gets simpler just a little bit.