In general, no. Consider an infinitely stretched dihydrogen molecule, in its ground (singlet) state. Obviously the HOMO is the in-phase linear combination of the 1s orbitals of the neutral hydrogen atoms, and the LUMO is their out-of-phase linear combination. (The shape of the HOMO follows from the fact that the density of the system must be equal to a sum of the densities of two neutral hydrogen atoms, and the shape of the LUMO follows from the fact that the Kohn-Sham potential is local, thus the LUMO of this molecule must be locally identical to the HOMO up to a phase.) A corollary is that, the square of the LUMO is the same as the square of the HOMO. (EDIT: more rigorously speaking, their difference is non-zero for finitely stretched hydrogen molecules but vanishes in the infinitely stretched limit.)
Now excite the molecule to the lowest doubly excited singlet state. It follows that in this state, at least one of the electrons must populate an atomic orbital other than the 1s orbital (EDIT: a molecular orbital involving significant contribution from a non-1s atomic orbital), since there are only three singlet states constructible from the two 1s orbitals and two electrons (the ground state and two charge transfer states:
$$\frac{1}{\sqrt{2}}(\psi_{\rm H1}(r_1)\psi_{\rm H2}(r_2) + \psi_{\rm H1}(r_2)\psi_{\rm H2}(r_1)), \psi_{\rm H1}(r_1)\psi_{\rm H2}(r_1), \psi_{\rm H1}(r_2)\psi_{\rm H2}(r_2)$$
The fourth linear combination constructible from the four Slater determinants, $\frac{1}{\sqrt{2}}(\psi_{\rm H1}(r_1)\psi_{\rm H2}(r_2) - \psi_{\rm H1}(r_2)\psi_{\rm H2}(r_1))$, is a triplet state), none of which is a doubly excited state. Therefore the density of this excited state cannot be given as twice the square of the LUMO. QED.
The conclusion may seem a bit non-intuitive; many will immediately say "isn't the $({\rm HOMO})_2 \to ({\rm LUMO})_2$ excitation a double excitation?" This reveals a fundamental difference between the Kohn-Sham and HF descriptions of this system. In HF, the closed-shell ground state wavefunction of this system can be expanded in atomic orbitals (neglecting the spin part):
$$
\psi_{\rm HF} = \frac{1}{2}(\psi_{\rm H1}(r_1)\psi_{\rm H2}(r_2) + \psi_{\rm H1}(r_2)\psi_{\rm H2}(r_1) + \psi_{\rm H1}(r_1)\psi_{\rm H2}(r_1) + \psi_{\rm H1}(r_2)\psi_{\rm H2}(r_2)) \tag{1}
$$
The latter two terms are spurious intermixtures of charge transfer states, which are famously responsible for the size-inconsistency of RHF. In this case we can formally write the $({\rm LUMO})_2$ determinant
$$
\tilde{\psi}_{\rm HF} = \frac{1}{2}(\psi_{\rm H1}(r_1)\psi_{\rm H2}(r_2) + \psi_{\rm H1}(r_2)\psi_{\rm H2}(r_1) - \psi_{\rm H1}(r_1)\psi_{\rm H2}(r_1) - \psi_{\rm H1}(r_2)\psi_{\rm H2}(r_2)) \tag{2}
$$
But this is just as unphysical as the RHF description of the ground state, again due to the charge transfer state contributions.
The Kohn-Sham wavefunction can of course be expanded as Eq. (1) as well, but a more important thing is that, the ground state Kohn-Sham density must equal the exact density, given by the exact wavefunction:
$$
\psi_{\rm exact} = \frac{1}{\sqrt{2}}(\psi_{\rm H1}(r_1)\psi_{\rm H2}(r_2) + \psi_{\rm H1}(r_2)\psi_{\rm H2}(r_1)) \tag{3}
$$
while the exact wavefunction of the first doubly excited state inevitably involves the 2s and/or 2p orbitals (EDIT: inevitably involves molecular orbitals that have large contributions from the 2s and/or 2p orbitals), and cannot have any simple relation with the exact ground state wavefunction, as we do in the RHF case. Therefore, while $({\rm LUMO})_2$ is a valid KS determinant, it does not have any simple relation with the density of the first doubly excited state.
Finally, there is one more thing to consider: doubly excited states are not well-defined concepts, especially in complex molecules, because the doubly excited determinants may mix extensively with singly excited determinants, making it impossible to unambiguously assign some states as singly or doubly excited. Thus, in some cases your question may not even be well-defined.
