Let $\alpha$ be an irrational. We shall consider its continuted fraction $[a_0;a_1,a_2,\dots]$. Recall some basic results about convergents of continued fractions (see e.g. here): letting $p_n,q_n$ be the sequence of numerators and denominators of convergents, for any $n>1$ we have $q_{n+1}=a_{n+1}q_n+q_{n-1}>a_{n+1}q_n$ and $$\left|\alpha-\frac{p_n}{q_n}\right|<\frac{1}{q_nq_{n+1}}<\frac{1}{a_{n+1}q_n^2},$$ hence $q_n|q_n\alpha-p_n|<\frac{1}{a_{n+1}}$. Therefore if $a_{n+1}$ are unbounded, then $q_n|q_n\alpha-p_n|$ does not attain a minimum. Therefore $R(\alpha)$ does not exist for those $\alpha$.
Therefore $R(\alpha)$ can only exist for badly approximable numbers. It is known that those numbers form a set of measure zero, and most natural constants besides quadratic irrationalities, including $\pi$ and all higher degree algebraic irrationals, are conjectured to not lie in it. Therefore $R(\pi)$ probably doesn't exist.
On the other hand, if $\alpha$ is badly approximable, then this still doesn't necessarily mean $R(\alpha)$ necessarily exists, as you note with $\alpha=\frac{1+\sqrt{5}}{2}$. In fact I believe it won't exist for any quadratic irrational. However, using the bound $$\left|\alpha-\frac{p_n}{q_n}\right|>\frac{1}{q_n(q_{n+1}+q_n)}>\frac{1}{(a_{n+1}+2)q_n^2},$$ we at the very least get that for those numbers the quantity $q_n|q_n\alpha-p_n|$ is bounded away from zero (note that $q|q\alpha-p|$ can only be smaller than $1/2$ if $p/q$ is a convergent, so we don't lose much from looking at just looking at convergents).
Last remark I have is that there are uncountably many $\alpha$ for which $R(\alpha)$ exists. Indeed, from the above considerations it follows easily that this is the case if some some $N$ we have that the continued fraction of $\alpha$ contains a partial denominator $N$, but from some point on all denominators are at most $N-2$.