 Solution

No. Take the Hausdorff space $X=\mathbb{R}$ and declare all nonzero points equivalent, leaving $0$ in its own class. The quotient has two points,The nonzero class is open because its inverse image $\mathbb{R}\setminus \{0\}$ is open. However, the only saturated set containing $0$ but not the other class is $\{0\}$, which is not open. Thus every neighbourhood of $[0]$ is the whole quotient, so the two quotient points cannot have disjoint neighbourhoods. Therefore $\tilde{X}$ need not be Hausdorff.