This is not a full answer, but firstly for 1, yes, of course. What's more, the elements of $R^+$ are independent of each other, because the elements of $R$ are independent of each other, and the pseudo-inverse preserves matrix rank.

For 3, $E[R^+_{ij}]=0$, due to symmetry, and $E[(R^+_{ij})^2]=\frac{1}{d(d-k-1)}$, as can be shown from an analysis in [this paper][1]. If you look into that paper, keep in mind that the singular inverse-Wishart distributed matrix would be $R^{+\top}R^+$, so the diagonal elements of the mean of that matrix, given in section 4.1, are simply $k$ times the variance of each element, i.e. $E[R^{+\top}R^+]=kE[(R^+_{ij})^2]I_d$. This is because the diagonal elements in the matrix product $R^{+\top}R^+$ each sum $k$ squared (independent) elements of $R^+$. Plugging in the derived equation from the paper, which in this post's notation would read $E[R^{+\top}R^+]=\frac{k}{d(d-k-1)}I_d$, gives us our answer. One could also get here via a similar path using the mean of the [non-singular Wishart distributed][2] random matrix, $E[R^+R^{+\top}]$.

As for 2, while each element of $R^+$ is generally not Gaussian (consider k=d=1), I would certainly believe they approach Gaussianity as the dimensions increase, but I don't know anything about that.


  [1]: https://projecteuclid.org/journals/electronic-journal-of-statistics/volume-5/issue-none/On-the-mean-and-variance-of-the-generalized-inverse-of/10.1214/11-EJS602.full
  [2]: https://en.wikipedia.org/wiki/Inverse-Wishart_distribution