Let $M$ be a Riemannian manifold, $x$ and $y$ are two points in $M$. 
Assume that $x$ is not in the cut locus of $y$. Does there exist a neighborhood $U$ of $x$ and a neighborhood $V$ of $y$ such that for every point $u$ in $U$ and for every point $v$ in $V$ we have that $u$ is not in the cut locus of $v$?



An elementary question about the cut locus
For a unit tangent vector $u$ with footpoint $p$ let $t(u)$ be the supremum of positive numbers such that the geodesic $t\to \exp\_p(tu)$ is minimizing on $[0,t(u)]$. The  *cut locus* at $p$ is the set of points $\exp\_{p}(t(u) u)$ of $M$ for which $t\_u$ is finite. 


A basic result is that $u\to t(u)$ defines a continuous map from the unit tangent bundle to $(0,+\infty]$ where continuity at $+\infty$ is understood in the obvious way. See e.g. Sakai's "Riemannian geometry", Proposition III.4.1.


Now coming to your question fix $x\in M$ and $y=\exp\_x(su)$ for some $u=u(x,y)$ and positive number $s$. Suppose $x^\prime$, $y^\prime$ are near $x$, $y$ respectively,
and write $y^\prime=\exp\_{x^\prime}(s^\prime u^\prime)$.
If $y$ is not in the cut locus of $x$, then $t(u)=+\infty$. So $t(v)> s$ on some neighborhood of $u$ in the unit tangent bundle. Since $s^\prime$ and $s$ are almost the same, we conclude that $t(u^\prime)> s^\prime$, i.e. $x^\prime$, $y^\prime$ are not cut points of each other.


In my view Sakai's book is the best source of information about cut and cunjugate loci.


