Since $H^2(\mathfrak{g},\mathbb{R})=0$ and since $G$ is simply connected, it follows from the van Est theorems that $H^2(G,S^1)=0,$ which means that all $S^1$- extensions of $G$ are trivial. But a lifting of a projective representation $\rho$ of $G$ is equivalent to the data of a trivialization of the $S^1$ - extension of $G$ given by the pullback (by $\rho$) of the extension $1\to S^1\to U(\mathcal{H})\to \mathbb{P}(U(\mathcal{H}))\to 1,$ which from the above van Est argument must exist.

For the relevant van Est theorems, see https://mathscinet.ams.org/mathscinet-getitem?mr=59285
or Theorem 2.3 in
https://arxiv.org/abs/1909.12100

Edit: In the finite dimensional case the Lie algebra extension associated to the extension $1\to S^1\to U(\mathcal{H})\to \mathbb{P}(U(\mathcal{H}))\to 1$ splits, and the proof I gave can be tweaked to show that projective unitary representations of simply connected Lie groups always lift to unitary representations. It probably doesn't split in the case that the hilbert space is infinite dimensional and this is why the assumption that $H^2(\mathfrak{g},\mathbb{R})=0$ is needed. In this case there is a slight gap in my proof since I didn't show that the pullback of the extension is actually a smooth $S^1$ - extension of $G$, but maybe this isn't hard to show.