I can make you a directed model! Let's specify the out-degree of each vertex: $o_x$ is the out-degree of vertex $x$.

Choose yourself an increasing function $F(d)$ that encodes a distance penalty. For each pair of vertices $x$ and $y$, let $A_{xy}=F(\|x-y\|)U_{xy}$. The edges leaving vertex $x$ are then the $\vec{xy}$ corresponding to the $o_x$ smallest values of $A_{xy}$.

Doing the undirected case seems a bit more subtle...

EDIT: Let me expand/change this a bit. For a mathematically interesting model, where you *might* be able to prove something, you could look at a Gibbs probability distribution. You define an "energy" for each legal configuration (i.e. subgraph of $G$ satisfying the degree constraints). Then, based on the hypothesis that high energy states are unlikely, you assign them low probability.

More specifically, a reasonable approach would be to define the energy of a configuration $\xi$ to be $\Phi(\xi)=\sum_{e\in E(\xi)}F(\|e\|)$. If you let $\Lambda$ be the set of all legal configurations, then the Gibbs measure is defined by $\mathbb P(\xi)=e^{-\Phi(\xi)}/\sum_{\zeta\in\Lambda}e^{-\Phi(\zeta)}$.
(The normalization, which is sometimes called the partition function, $Z(\Lambda)$, there makes this a probability measure). The reason that Gibbs measures are nice is that the multiplicative properties of the exponential function ($e^{a+b}=e^ae^b$) lead to some independence properties of the measure you've constructed. For example it's easy to see that if $\xi$ and $\xi'$ are 2 configurations that agree except that $\xi$ contains edges $ab$ and $cd$, while $\xi'$ contains edges $ac$ and $bd$, there's an easily calculated relationship between $\mathbb P(\xi)$ and $\mathbb P(\xi')$.