$\Omega G$ won't be a differentiable stack unless you are willing to go to infinite dimensions. Provided you are considering $S^1$ as a smooth manifold, Nerses answer above is correct- it gives you a nice sheaf on the site of manifolds (it's even concrete- a so-called diffeological space). It's also representable by an infinite dimensional manifold.

The other thing you could mean, is you might mean $S^1$ (or $B \mathbb{Z}$) as a homotopy type, or the "categorical circle." For this, you will have to go to derived manifolds (otherwise you will just recover $G$ back again), and then, $\Omega X = T[1] X$ is the shifted tangent bundle for any manifold $X,$ in particular, for $X=G$. $T[1] X$ is the graded manifold whose underlying space is $X$ and whose structure sheaf $\mathcal{O}$ is given by $\mathcal{O}^{-n}(U)=\Omega^{n}_{dR}(U).$

*Edit*: Actually, in the above, I am computing $\mathcal{L} X$ the *free loops* on $X$:

For $x \in X,$ the based loops at $X$ are the homotopy fibered product $* \times_{X} T[1]X$, but since evaluation at the base point of $S^1$ is just the vector bundle map $T[1]X \to X,$ which is a submersion, the ordinary pullback is a homotopy pullback, so we get simply $\Omega_x X=T_x[1] X=\left(T_x X\right)[1],$- the graded manifold associated to the graded vector space with $T_x X$ sitting in degree $-1.$ It's underlying manifold is just a point, but it's structure sheaf is given by the exterior algebra of the dual this vector space (but placed in non-positive degrees).