Is the (super-)symmetric power of the exterior algebra free?

Let $V$ be a vector space over $k$ of dimension $m$. (I'm only interested in the case $k=\mathbb{Q}$.) Let $R:=\Lambda^*V$ be the exterior algebra. It carries the structure of a supercommutative ring: $R=R^+\oplus R^{-}$, where both the odd and the even part have $k$-dimension $2^{m-1}$. We define a multiplication on $R^{\otimes n}$ by $(\alpha_1\otimes\cdots\otimes\alpha_n)\cdot (\beta_1\otimes\cdots\otimes\beta_n) =\pm (\alpha_1\beta_1\otimes\cdots\alpha_n\beta_n)$, where the sign comes from reordering factors of odd degree. The symmetric group $\mathfrak{S}_n$ acts on $R^{\otimes n}$ by permuting the factors such that a transposition of two factors of odd degrees yields a negative sign.

The supersymmetric power $S^nR$ is defined as the quotient of $R^{\otimes n}$ by this action. Over $k=\mathbb{Q}$ this is the same as taking the subspace of invariants. The canonical inclusion of $V$ in $S^nR$ gives $S^nR$ the structure of a $R$-algebra.

Now I conjecture that $S^nR$ is free as a module over $R$ for $n\geq 1$.

Let me give evidence to this. As a $k$-vector space, $S^nR$ can be decomposed as: $$S^nR = \bigoplus_{p+q=n} \text{Sym}^p(R^+)\otimes \Lambda^q(R^{-}).$$ The generating function of the dimensions of $S^nR$ is therefore $\sum_{n\geq 0} \dim_k(S^nR)\, t^n= \left(\frac{1+t}{1-t}\right)^{2^{m-1}},$ which can be seen to have all coefficients divisible by $2^m=\dim_k R$ except for the first one.

For instance, if $V$ is one-dimensional, then $R$ is isomorphic to $k[x]/(x^2)$ and $S^nR = R$ for all $n\geq1$. If $V$ is two-dimensional with basis $\{\theta_1,\theta_2\}$, then $S^2R$ is generated over $R$ by $1\otimes 1$ and $\theta_1\otimes \theta_2$.

Any idea of proof or references will be gratefully welcomed.

I found a proof for the desired freeness result in the case $\text{char}(k)=0$, but it is a bit technical.
The idea is to show first, that $R^{\otimes n}$ is a free $R$-module in a way that the action of $\mathfrak S_n$ is $R$-linear. The second step is to find a basis of $S^n R$, when the basis of $R^{\otimes n}$ already is constructed.
For some $v\in V$ we denote $v^{(i)} := 1^{\otimes i-1}\otimes v\otimes 1^{\otimes n-i+1} \in R^{\otimes n}$. Then $R^{\otimes n}$ is generated as a $k$-algebra by the elements $\{v_j^{(i)}\}$ for a $k$-basis $\{v_j\}$ of $V$. Now consider the ring automorphism $$\sigma : R^{\otimes n} \longrightarrow R^{\otimes n}, \\ v^{(1)} \longmapsto v^{(1)} +v^{(2)} + \ldots + v^{(n)}, \quad v^{(i)} \longmapsto v^{(i)} \text{ for } i>1.$$ Now look at the injection $$\iota: R \rightarrow R^{\otimes n},\quad \theta \mapsto \theta\otimes 1\otimes\cdots\otimes 1.$$ It is clear that this turns $R^{\otimes n}$ into a $R$-algebra and in particular into a free $R$-module. Denote $\{b_i\}$ a $R$-linear basis of $R^{\otimes n}$. But $R^{\otimes n}$ can also be considered as a free $R$-module with the $R$-module structure given by $\sigma\iota : R \rightarrow R^{\otimes n}$.
The image of $\sigma\iota$ is fixed by the action of $\mathfrak S_n$. Thus the kernel and the image of any $\pi\in \mathfrak S_n$ become free $R$-modules as well. In particular, the space of invariants, given by the image of $\frac{1}{n!}\sum\limits_{\pi\in\mathfrak S_n}\pi$ which is isomorphic to $S^nR$, is a free $R$-module.