There are a number of technical issues with making what you describe precise, for example: what precisely is a supermanifold with boundary? how can you glue/compose bordisms? etc. Many or all of these have been or can be surmounted, and I would suggest looking at the work of Stolz-Teichner and the reference therein to see what sort of things people have tried to do.

In any case there is the fact that (in the smooth category) every $(d|k)$-dimensional supermanifold is (non-canonically!) of the form $\pi E$ for $E$ a rank $k$ vector bundle over a $d$-manifold.

Thus, unless you do something more fancy like work in families, when you pass to bordism classes you will just get the bordism group of $d$-manifolds equipped with rank $k$-vector bundles. This has two names:

$$ \pi_d( MO \wedge BO(k)) = MO_d(BO(k))$$

either as the d-th homotopy group of the smash of the spectrum $MO$ and space $BO(k)$ (which is also another Thom spectrum) or equivalently as the d-th $MO$-homology group of $BO(k)$. So that identifies the bordism group.

I think it is a very interesting question whether the whole Pontryagin-Thom construction can be carried out inside the world of supermanifolds, but that is a different question.

There are a number of technical issues with making what you describe precise, for example: what precisely is a supermanifold with boundary? how can you glue/compose bordisms? etc. I am going to ignore these technicalities because I don't think it makes much of a difference to your question. Many of these technical issues have been addressed by other people, for example I would suggest looking at the work of Stolz-Teichner and the reference therein to see what sort of things people have tried to do.

In any case in the smooth category we have:

Batchelor's theorem: every $(d|k)$-dimensional supermanifold is (non-canonically!) of the form $\pi E$ for $E$ a rank $k$ vector bundle over a $d$-manifold. Moreover the isomorphism class of the vector bundle is uniquely determined by the super manifold.

Here $\pi E$ is the super manifold whose ring of functions is the global sections of the exterior algebra bundle $\wedge^* E^*$. So the morphisms from $\pi E$ to $\pi E'$ come from all algebra maps and are more than just the vector bundle maps (which correspond to homogeneous algebra maps).

If $(X, \mathcal{O}_X)$ is a super manifold, the vector bundle $E$ can be obtained by considering $\mathcal N / \mathcal N^2$ where $\mathcal N$ is the subsheaf of $\mathcal{O}_X$ of nilpotent elements.

So there are functors in both directions but there is no natural isomorphism from the identity functor on supermanifolds to $\pi(\mathcal N / \mathcal N^2)$. This is what the "non-canonical" means.

But we still get a bijection between isomorphism classes of supermanifolds and isomorphism classes of manifolds with vector bundles. Thus, unless you do something more fancy like work in families, when you pass to bordism classes you will just get the bordism group of $d$-manifolds equipped with rank $k$-vector bundles. This has two names:

$$ \pi_d( MO \wedge BO(k)) = MO_d(BO(k))$$

either as the d-th homotopy group of the smash of the spectrum $MO$ and space $BO(k)$ (which is also another Thom spectrum) or equivalently as the d-th $MO$-homology group of $BO(k)$. So that identifies the bordism group.

I think it is a very interesting question whether the whole Pontryagin-Thom construction can be carried out inside the world of supermanifolds, but that is a different question.