Since you used the word "orthogonal" I assume you are working on a Riemannian manifold. So let $(M,g)$ be Riemannian, and $X$ a vector field. Your question actually has several subquestions buried in it. Let me address each one in turn.

### Orthogonal hypersurface

Whether there exists (even locally) a hypersurface that is orthogonal to $X$ is a well-studied problem, and the local problem is completely solved by Frobenius's Theorem. An equivalent formulation to your question is, if we let $\omega$ be the metric dual one form to $X$ (so $\omega(Y) = g(X,Y)$), whether the kernel of $\omega$ is tangent to a hypersurface. The local obstruction is given by $\omega\wedge (d\omega)$; locally such an hypersurface exists if and only if this quantity vanishes.

### Flow mapping these slices

Obviously this cannot hold in general. If $X$ is a vector field and $Y = fX$ for some scalar non-vanishing function $f$, and $X$ is hypersurface orthogonal, then obviously so is $Y$. But the flows defined by $X$ and $Y$ are in general completely different, so at most one of the two can preserve orthogonality.

If we take a step back and forget about the integrability problem, we see that you can formulate this at the infinitesimal level by asking that the flow map preserving the orthogonal complement to $X$. In terms of the dual one form $\omega$, this is
$$ \mathcal{L}_X(\omega) \wedge \omega = 0 $$
Using Cartan's magic formula this expands to
$$ d(\omega(X)) \wedge \omega + i_X(d\omega \wedge \omega) = 0 $$
so we see that when you already know that $\omega$ is hypersurface orthogonal, the flow maps these hypersurface to each other if and only if
$$ \nabla (g(X,X)) \propto X $$

### When $X$ is Killing

In this case the second result above becomes guaranteed, since $\mathcal{L}_X \omega = 0$. So the issue is only integrability. Not all Killing fields are hypersurface orthogonal. Write $\mathbb{R}^3$ in cylindrical coordinates $(r,z,\theta)$, then the vector field $\partial_z + \partial_\theta$ is Killing and nowhere vanishing. Its metric dual one form is $dz + r^2 d\theta$, whose exterior derivative is $2 r dr \wedge d\theta$, and so we see that hypersurface orthogonality fails.