Let $\mathbf{X}$ be an (orthogonal) spectrum (can assume that it's an $\Omega$-spectrum if this helps give a positive answer) and give the category of orthogonal spectra the stable model structure.
Let $\Omega^\infty$ be the total right derived functor of evaluation at $0$.
I want to show that $$ \Omega^\infty (S^n \wedge \mathbf{X}) \simeq \mathbf{X}(\Bbb{R}^n). $$
Attempt:
Since we can assume that $\mathbf{X}$ is an $\Omega$-spectrum, (hence fibrant in the stable model structure) $\Omega^\infty$ is simply just evaluation at $0$. So the left-hand side becomes $$ S^n \wedge \mathbf{X}(0). $$
This comes with an obvious structure map to $\mathbf{X}(\Bbb{R}^n)$, but this map, in general, is not a weak homotopy equivalence.