It is true under the assumption $k,d\to+\infty$ while $k/d\to 0$, in the sense that there exists a matrix $G\in R^{d\times k}$ with iid $N(0,1)$ entries such that the pseudo inverse $R^+$ satisfies
$$
\|R^+  - G\|_{op}
/\|G\|_{op} \to^P 0.
$$
Above, the denominator $\|G\|_{op}$ could be replaced by $\sqrt d$
since $\|G\|_{op}/\sqrt{d}\to^P 1$ because, e.g., of some well known concentration inequalities for the smallest and largest singular values of $G$, namely
$$
P( \sqrt{d} - \sqrt{k} - t \le s_{\min}(G) \le s_{\max}(G) = \|G\|_{op} \le \sqrt d + \sqrt k + t)
\le 2e^{-t^2/2}.
$$
(One can for instance use $t=\sqrt{\log(d)}$ to obtain vanising probabilities).

Let us now explain why $\|R^+  - G\|_{op}
/\sqrt{d} \to^P 0$ holds.

If $R$ has iid entries with $k$ rows and $d$ columns ($k<d$), then by rotational invariance the SVD $R=UDV^T$ satisfies 

- $U\in O(k)$ is uniformly distributed (Haar measure on $O(k)$,
- the diagonal matrix $D\in R^k$ contains the random singularvalues,
-  $V\in R^{d\times k}$ has $k$ orthonormal columns and is distributed according to the Grassmanian. 

And $(U, D, V)$ are independent.

The pseudo-inverse is $R^+ = VD^{-1} U^T$. Now, let $\tilde D$ be an independent copy of $D$, independent of everything else mentioned so far. Then $G=V\tilde D U^T$ is equal in distribution to a matrix with iid $N(0,1)$ entries because $V, \tilde D, U$ are independent and as in the bullet points above.
Because of the concentration of the smallest and largest eigenvalues,
$\|D^{-1}/\sqrt d -I_k\|_{op}\to^P0$ and $\|\tilde D/\sqrt d - I_k\|_{op}\to^P 0$ so that $\|D^{-1} - \tilde D\|_{op}/\sqrt{d} \to^P 0$. It follows that $\|G-R^+\|_{op}/\sqrt d=\|V(D^{-1} - \tilde D)U^T\|_{op}/\sqrt d \to^P 0$ as claimed.