Yes, there is a stably-framed bordism category. Recall that tangential structures on smooth $n$-manifolds can be parameterized by maps $X \to BO(n)$; an $X$-structure on $M$ is then by definition a lift of the classifying map for the tangent bundle $T_M : M \to BO(n)$ through $X$. In the stably-framed case, you take $X$ to be the fibre of $BO(n) \to BO = BO(\infty)$, equivalently $X = O/O(n)$. The cobordism hypothesis says what it always says: to give a fully-extended TQFT on manifolds with $X$-structure is to give a fully-dualizable object together with a trivialization of the $\Omega X$-action thereon. Remark: although people regularly say things like "oriented TQFT" and "framed TQFT" and the like, it would be more accurate to call the TQFT "cooriented" or "coframed", since it is the manifold which is oriented or framed — that way you can remember that a framing, say, is more data on the manifold, but a coframing is less data on the TQFT (since it needs to be defined on fewer manifolds).

The fibre of $BO(n) \to BO(n+1)$ is an $S^n$, and so this $X$ is filtered as $X = \dots.S^{n+2}.S^{n+1}.S^n$, i.e. it has one cell in each dimension $\geq n$. Thus to give a stable coframing to an $n$-dualizable object is to give one $n$-morphism, one $(n+1)$-morphism, one $(n+2)$-morphism, and so on. If your target $(\infty,n)$-category is just an $n$-category, then the $n$-morphism is data, and the $(n+1)$-morphism is an equation your data must solve, and all the higher cells are vacuous. The upshot is that many TQFTs are naturally co-framed.