Edited Jan31, 2022: the cut locus of the curve $\gamma$ is a finite tree if the boundary is smooth and if the curve bounds a contractible domain, i.e. $\Omega \approx \{pt\}$. This follows from Blum's MAT and the basic ideas of the theory. The observation about $C$ being a locally finite tree requires the boundary be smooth or $C^1$. In this case the ``branches" of the tree cannot accumulate, and the principle curvatures of the boundary hypersurface are bounded. Otherwise if the boundary is only $C^0$, then the cut locus $C$ might be a tree with infinite degree vertices.
N.B. It does not really make sense to speak of trees in this category, so terms like "branches" and "tree" are equally informal.
Let $A\subset \mathbb{R}^n$ be a bounded open subset. The cut locus $C$ of $A$ is defined as the domain of nondifferentiability of $dist_{\partial A}: A\to \mathbb{R}$.
Blum's Medial Axis Transform (MAT) is the set $$M(A):=\{x\in A ~|~ card|argmin_{y\in \partial A} dist(x,y)|\geq 2\}.$$
We find $M(A) \subset C$, although $M(A)$ is not always closed, with $\overline{M(A)}=C$.
(1) To answer the question posed in the OP's title: Both the cut locus $C$ and $M(A)$ have the homotopy type of the domain $A$, so if $A$ (or $\Omega$ in the notation of the OP) is contractible, then $C$ and $M(A)$ is contractible.
(2) To answer the OPs specific question about pathologies in $C$ and $M(A)$: the cut locus $C$ branches wherever the principal curvatures of the boundary $\partial A$ are positive (convex) with respect to the interior $A$. Saul Rodrigues Martin gives interesting example in Is the max-centre map continuous for open bounded domains?
My original answer to the OP was not satisfactory. But essentially I would argue that the cut locus, if interpreted as a type of tree graph, has finitely many graph-edges if the boundary $\gamma$ is basically $C^1$. But once you have infinitely many edges, it's possible to get those edges to accumulate or to share a common intersection, i.e. a vertex of infinite degree.