Suppose $\rho_1$ and $\rho_2$ are two Gaussian states. The trace distance of a Gaussian state with covariance matrix $V$ can be computed as the sum of the eigenvalues of $i\Omega V$, so to find the distance of two Gaussian states we need the covariance matrix of $\rho_1 - \rho_2$.
Does this have a simple expression in terms of the individual covariance matrices $V_1$ and $V_2$?
There are many things being overlapped here that likely have no business together. A Gaussian state is uniquely determined by its covariance matrix and its displacement vector, any linear combination of states has some covariance matrix and displacement vector, but the difference between two states is not uniquely defined by the covariance matrix and displacement vector of the latter. And lastly, it does not make sense to refer to "the trace distance of a Gaussian state" because distances must be between two objects. You can see that Mark Wilde asked this question from the intuition that the trace distance between two Gaussian states will not be something trivial. Or read this paper that says " the trace distance of two Gaussian states is not a handy object" and where the authors only go on to provide a bound for the distance in terms of fidelities.
With that all being said, we can answer some of the actual questions asked here, even if the premise does not apply. The displacement vector and covariance matrix of any object $\rho_1-\rho_2$ can be found from covariance rules, so long as operator ordering is respected (e.g., if you use symmetrix covariances $\mathrm{Cov}(X,Y)=\langle XY+YX\rangle/2-\langle X\rangle\langle Y\rangle$ then you can use the classical rules with no caveats): $$\langle R_i \rangle_{\rho_1-\rho_2}=\mathrm{Tr}(R_i(\rho_1-\rho_2))=\langle R_i\rangle_{\rho_1}-\langle R_i\rangle_{\rho_2}$$ and $$\langle R_i R_j\rangle_{\rho_1-\rho_2}=\langle R_i R_j\rangle_{\rho_1}-\langle R_i R_j\rangle_{\rho_2}.$$ I.e., means for differences between states are the differences between the means for each state. The displacement vector for the difference state is the difference between the displacement vectors for each state, while the covariance matrix is a tiny bit more intricate: \begin{align}\mathrm{Cov}_{\rho_1-\rho_2}(R_i,R_j)&=\langle R_i R_j+R_jR_i\rangle_{\rho_1}-\langle R_i R_j+R_jR_i\rangle_{\rho_2}-(\langle R_i \rangle_{\rho_1}-\langle R_i\rangle_{\rho_2})(\langle R_j \rangle_{\rho_1}-\langle R_j\rangle_{\rho_2})\\ &=\mathrm{Cov}_{\rho_1}(R_i,R_j)-\mathrm{Cov}_{\rho_2}(R_i,R_j)-2\langle R_i\rangle_{\rho_2}\langle R_j\rangle_{\rho_2} + \langle R_i\rangle_{\rho_1}\langle R_j\rangle_{\rho_2}+ \langle R_j\rangle_{\rho_1}\langle R_i\rangle_{\rho_2}.\end{align}
Therefore, if the individual covariance matrices are $\mathbf{V}_i=\mathrm{Cov}_{\rho_i}(\mathbf{R},\mathbf{R})$ with components $(\mathbf{V}_i)_{jk}=\mathrm{Cov}_{\rho_i}(R_j,R_k)$ and the displacement vectors are $\mathbf{d}_i=\langle\mathbf{R}\rangle_{\rho_i}$ then $$\mathbf{V}_{\rho_1-\rho_2}=\mathbf{V}_1-\mathbf{V}_2-2\mathbf{d}_2\mathbf{d}_2^\top+\mathbf{d}_1\mathbf{d}_2^\top+\mathbf{d}_2\mathbf{d}_1^\top,$$ which of course simplifies for zero-mean states with $\mathbf{d}=\mathbf{0}$.
Again, I repeat that $\rho_1-\rho_2$ is not a Gaussian state and so its displacement vector and covariance matrix are not the only information one needs for describing it, but this is the answer to the question as posed.
