Is it true that any totally geodesic hypersurface in a nonpositively curved manifold has a tubular neighborhood such that the metric on the neighborhood is a warped product? At least, if the manifold is simply connected?
The paper "Lower bounds on Ricci curvature and almost rigidity of warped products" by Cheeger and Colding might be useful to you. In short, they proved that a locally warped product piece of a manifold must satisfy $Hess(F)=kg$ for some functions $F$ and $k$, where $k=F''$. (See page 192. We can also see that $F'$ is the warping function.) And this condition is actually sufficient.
It seems to me that your totally geodesic hypersurface, say $S$, must be a level set of $F$. In this case, we have $Hess(F)=0$ because the second fundamental form of $S$ is $Hess(F)/|\nabla F|$. This implies that $k=0$ (otherwise we have $g=0$, it's impossible.) In this case, you can find that $F$ is a linear function in the radial direction and in fact you get a (non-warped) product manifold. So your hope could be true only if your manifold could be of product type at least. (Then by Anton's comment, it is indeed rigid.)
On the other hand, I don't think the condition "simply-connected" is crucial because you can always assume it by passing to the universal cover.