The following answer to Question 1 is based on Alain Valette's response, it is too long for a comment.

For $X\in Z(\mathcal{U}\mathfrak{g})$ the actions of $X$ and $G^0=\mathrm{GL}^+_n(\mathbb{R})$ commute, hence we are done by Schur's lemma when $\pi|G^0$ is irreducible. Otherwise $\pi|G^0$ is the sum of two irreducible representations $(\rho^\pm,H^\pm)$ which are conjugates of each other by the matrix $i:=\mathrm{diag}(-1,1,\dots,1)\in P$. In particular, $i$ interchanges $H^\pm$. By Schur's lemma, $X$ acts on $H^{\pm}$ by some scalars $c^\pm(X)\in\mathbb{C}$. We are done if $c^+(X)=c^-(X)$.

Assume that $c^+(X)\neq c^-(X)$, and let $f\in C_0^\infty(G)$ be supported in a connected component of $G$. As $R$ commutes with $\pi(i)$, we have
$$\mathrm{trace}(R\pi(Xf))=\mathrm{trace}(R\pi(i)\pi(Xf)\pi(i)).$$
Note that $\pi(f)$ either leaves the spaces $H^\pm$ invariant or interchanges them, depending on whether $f$ is supported in $G^0$ or in $G^0i$. In either case the previous equation 
together with
$$\Lambda_R(f)=\mathrm{trace}(R\pi(f))=\mathrm{trace}(R\pi(i)\pi(f)\pi(i))$$
yields
$$c^+(X)\Lambda_R(f)=c^-(X)\Lambda_R(f).$$
This shows that $\Lambda_R(f)=0$, hence $\Lambda_R=0$ as a whole.