Let $\pi:(\tilde{M},\tilde{g})\to (\mathbb{R},g)$ be a Riemannian submersion. Given a vector field $X$ on $\tilde{M}$ which at each point $p\in\tilde{M}$ is unitary and orthogonal to $\pi^{-1}(q)$, where $\pi(p)=q.$ Now if one can guarantee that $X$ is projectable?
P.S. $X$ is projectable if for fixed $q\in \mathbb{R}$, $d\pi(X_p)$ is independent of $p\in \pi^{-1}(q)$.
Also note that each fiber is of codimension 1.
The map $$f :\pi^{-1}(q) \to T_q \mathbb R : f(p) = d\pi(X_p)$$ is continuous, and by the assumptions ($X_p \in (\ker d \pi)^\perp$, $|X_p|=1$ and the fact that $\pi$ is a Riemannian submersion) we know that $|f(p)| = 1$. Thus $f$ in fact takes values in the $0$-sphere $\{ -1, +1 \} \subset T_q \mathbb R$; so assuming the fibers are connected we conclude that $f$ is constant; i.e. $X$ is projectable.
If you allow disconnected fibers then there are obvious counterexamples - e.g. just disjoint union an example with a copy of itself where $X$ is flipped.