Note that if $T\to S$ is also quasicompact, then $S$ must be locally noetherian: this boils down to the well-known fact that if $A\to B$ is a faithfully flat ring homomorphism and $B$ is noetherian, then so is $A$.

This proves that Jason's example above is indeed a counterexample. More generally, any non-noetherian local scheme $S$ is a counterexample: if $T\to S$ is faithfully flat, there is an open affine $U\subset T$ which covers $S$.

Note that if $T\to S$ is also quasicompact, then $S$ must be locally noetherian: this boils down to the well-known fact that if $A\to B$ is a faithfully flat ring homomorphism and $B$ is noetherian, then so is $A$.

This proves that Jason's example above is indeed a counterexample. More generally, any non-noetherian local scheme $S$ is a counterexample: if $T\to S$ is faithfully flat, there is an open affine $U\subset T$ which covers $S$.

EDIT: In fact, here is a complete answer ($S$ is any given scheme):

(1) The following are equivalent:
(1a) There exists a locally noetherian scheme $T$ and a faithfully flat and quasicompact morphism $T\to S$.
(1b) $S$ is locally noetherian.

(2) The following are equivalent:
(2a) There exists a locally noetherian scheme $T$ and a faithfully flat morphism $T\to S$.
(2b) For each $s\in S$, the ring $\mathcal{O}_{S,s}$ is noetherian.

Proof: exercise. To show that (2b) implies (2a), take $T=\coprod_{s\in S}\mathrm{Spec}(\mathcal{O}_{S,s})$.