It is true under the assumption $k,d\to+\infty$ while $k/d\to 0$, in the sense that $R^T\in R^{d\times k}$ has obviously iid $N(0,1)$ entries, and satisfies
$$
\|\sqrt d R^+  - R^T/\sqrt d\|_{op}
\to^P 0.
$$
First let us recall some well known concentration inequalities for the smallest and largest singular values of $R$, namely
$$
P( \sqrt{d} - \sqrt{k} - t \le s_{\min}(R) \le s_{\max}(R) = \|R\|_{op} \le \sqrt d + \sqrt k + t)
\le 2e^{-t^2/2}.
$$
if $k/d\to 0$, one can for instance use $t=\sqrt{\log(d)}$ to obtain  that $s_{\min}(R)/\sqrt d\to^P 1$ and similarly $\|R\|_{op}/\sqrt d\to ^P 1$.

Let us now explain why $\|\sqrt d R^+  - R^T/\sqrt d\|_{op}
\to^P 0$. Consider the SVD $R=UDV^T$. Then the pseudo-inverse is
$R^+ = VD^{=1} U^T$ and
\begin{align*}
\|\sqrt d R^+  - R^T/\sqrt d\|_{op}
&=\|U(\sqrt d D^{-1} - d^{-1/2} D)V\|_{op}
\\&=\|\sqrt d D^{-1} - d^{-1/2} D\|_{op}
\\&\le
\|\sqrt d D^{-1} -I_k\|_{op}
+
\|I_k - d^{-1/2} D\|_{op}.
\end{align*}
If $k/d\to 0$, the right-hand side converges to 0 in an event of probability at least $1-2/p$ by taking $t=\sqrt{\log d}$ in the above concentration inequality.