Here is an answer of sorts.

For simplicity, I restrict to cocompact torsion-free lattices. Then $b_{odd}(X_\Gamma)=0$ implies that $X$ is even-dimensional. The situation is clear if $dim(X)\le 2$, so the first interesting case is when $dim(X)=4$. There are (essentially) three 4-dimensional symmetric spaces of noncompact type (up to rescaling the metric on de Rham factors): The real-hyperbolic space, ${\mathbb H}^4$ (when $G=PO(4,1)$), the complex-hyperbolic space ${\mathbb H}^2_{\mathbb C}$ (when $G=PU(2,1)$) and the product of two hyperbolic planes ${\mathbb H}^2\times {\mathbb H}^2$ (when $G=PO(2,1)\times PO(2,1)$), up to commensuration of Lie groups.

$X={\mathbb H}^4$. I do not know if there are examples with $b_1=0$. However, it is known that for every arithmetic uniform lattice $\Gamma< PO(4,1)$ there exists a congruence-subgroup $\Gamma_1<\Gamma$ such that $b_1(\Gamma_1)= b_1(X_{\Gamma_1})>0$ (Millson). I do not think anybody knows what happens with non-arithmetic manifolds.

$X={\mathbb H}^2_{\mathbb C}$. Arithmetic lattices come in two types: All lattices of type I again have congruence-subgroups $\Gamma_1$ with $b_1(\Gamma_1)>0$ (Kazhdan). However, for all congruence-subgroups $\Gamma_1$ in type II arithmetic lattice,
$b_1(\Gamma_1)=b_3(\Gamma_1)=0$. So all odd Betti numbers vanish. The question of positivity of 1st Betti number of finite-index non-congruence subgroups is wide-open. The non-arithmetic case is mostly open. (There is one nonvanishing result, due to Yeung.)

$X={\mathbb H}^2\times {\mathbb H}^2$. The case of reducible lattices is (essentially) clear, so I will consider irreducible lattices. For all such lattices $\Gamma$, $b_1(\Gamma)=b_3(\Gamma)=0$. (This follows from Margulis' super-rigidity, but might have been known earlier.) So all odd Betti numbers vanish.

I will not attempt to analyze the case when $dim(X)\ge 6$. One gets vanishing results for $b_1$, but the situation with higher odd Betti numbers are very much unclear, unless you know something about odd Betti numbers of the dual compact metric space: Nonvanishing of some of those implies nonvanishing of $b_{odd}(\Gamma)$. There are vanishing theorems going back to Matsushima, but they do not deal with **all** odd Betti numbers.