One can indeed modify the formulas in the paper of Getzler to get an answer for any $\mathbf R^d$, by judiciously inserting minus signs and making substitutions $x \mapsto x^{d-1}$ in various places, but if I try I'll probably get it wrong. So let me explain how it works instead, and hopefully there are no sign errors and everything ends up in the right cohomological degree.

The cohomology ring of $F(\mathbf R^d,k)$ was computed by Cohen. It is the graded commutative algebra generated by variables $\omega_{ij}$ of degree $(d-1)$ for distinct $i,j \in \{1,\ldots,k\}$, modulo the relations

- $\omega_{ij}^2 =0$,
- $\omega_{ij} = (-1)^d \omega_{ji}$,
- $\omega_{ij}\omega_{ik} + \omega_{jk}\omega_{ji} + \omega_{ki}\omega_{kj} = 0$.

One can associate a graph to a monomial in these generators, with vertices $\{1,\ldots,k\}$ and an edge between $i$ and $j$ if $\omega_{ij}$ occurs in the monomial. The relations imply that the monomial is zero unless this graph is a forest.

The top degree $H^{k(d-1)}(F(\mathbf R^d,k))$ is spanned by monomials such that the corresponding graph is a tree. One can interpret such a monomial as a bracketing of $\{1,\ldots,k\}$: if $d$ is odd then the second and third relation correspond to anticommutativity and Jacobi identity for this bracketing, so the $S_k$-representation is given by the space $\mathrm{Lie}(k)$ spanned by Lie words on $k$ letters, and if $d$ is even we get instead $\mathrm{Lie}(k) \otimes \mathrm{sgn}_k$.

In lower degrees we get instead a sum over partitions of $\{1,\ldots,k\}$, and for each block $B$ in the partition a monomial such that the corresponding graph is a tree on the vertex set $B$. This tells us that we are considering a plethysm of symmetric functions.

Let $\widehat\Lambda$ be the degree completion of the ring of symmetric functions in infinitely many variables and let $\widehat\Lambda[[t]]$ be the power series ring in one variable over it. If $V$ is a representation of $S_k$, let $\mathrm{ch}_k(V)$ be the corresponding degree $k$ symmetric function. Define
$$ C = \sum_{k \geq 0} h_k \in \widehat\Lambda,$$
the sum of the trivial representation in each arity. Let
$$ L_d = \begin{cases} \sum_{k \geq 0} \mathrm{ch}_k \mathrm{Lie}(k) \otimes \mathrm{sgn}_k t^{k(d-1)} \in \widehat \Lambda [[t]] & d \text{ even} \\ \sum_{k \geq 0} \mathrm{ch}_k \mathrm{Lie}(k) t^{k(d-1)} \in \widehat \Lambda [[t]] & d \text{ odd} \end{cases}$$
Then
$$ \sum_{k \geq 0} \sum_{i \geq 0} t^i \mathrm{ch}_k H^i(F(\mathbf R^d,k)) = C \circ L_d.$$
The plethysm here satisfies $p_n \circ f(t,p_1,p_2,\ldots) = f(t^n,p_n,p_{2n},\ldots)$. You can also write this plethysm in terms of an infinite product or the "plethystic exponential", like Getzler does. One can describe $\mathrm{ch}_k\mathrm{Lie}(k)$ explicitly in terms of symmetric functions, and Getzler is using this in his paper. Namely, there is an isomorphism $\mathrm{Lie}(k) = \mathrm{Ind}_{C_k}^{S_k} \chi$ where $\chi$ is a primitive character of the cyclic group on $k$ elements, and
$$ \mathrm{Ind}_{C_k}^{S_k} \chi = \frac 1 k \sum_{{d \mid k}} \mu(d)p_k^{k/d}.$$

Lurking in the background is an operadic perspective. The operad $e_d$ is given by the homology of the little $d$-dimensional disks and what we are trying to do is calculate the characteristic of this operad. The operad $e_d$ can be constructed as the product of the commutative operad and a $(d-1)$-fold suspension of the Lie operad by a distributive law. This tells us that the characteristic of $e_d$ is the plethysm of the characteristic of $\mathrm{Com}$ (which is $C$) and the characteristic of this $(d-1)$-fold suspension, which is $L_d$.