I would expect many of your graphs to not have distinct eigenvalues.
Here's one reason: They have a very large number of degree $1$ vertices. Enough, in fact, that there will probably be a decent number of degree $1$ vertices which share a common neighbor with another degree $1$ vertex. These will give rise to many independent vectors in the nullspace of your matrix, so $0$ will be a multiple eigenvalue.
More generally, you can imagine the following situation: We say a subgraph $G'$ "dangles" from a vertex $v$ if the only edge connecting $G'$ to the remainder of the graph goes from some vertex $w \in G'$ to $v$. If you have two copies of $G'$ dangling from the same $v$ in the same way ($w$ occupies the same position in both graphs), then all the eigenvalues of $G'$ appear in the spectrum of $G$: The eigenvector is equal to $x$ on one copy of $G'$, $-x$ on the other copy, and $0$ elsewhere in the graph, where $x$ is an eigenvector of $G'$.
You can check, for example, that if instead of a power law graph you had the Erdős-Rényi graph $G(n,c/n)$ for fixed $c$, then every tree $T$ of fixed size appears as a dangling subgraph $c'n$ times (where $c'$ is positive and depends on the tree and $c$). This in turn implies there will be on average $c''n$ vertices having two copies of $T$ dangling from them, and for every such vertex you'll get a copy of all the eigenvalues of $T$. You might expect a similar phenomenon to happen with power law graphs, though I haven't checked this. (I did check a couple of graphs from the Stanford collection to see if they had $0$ as a multiple eigenvalue, and the ones I checked had it many times over).
This is all dependent on there being a sizable number of degree one vertices though. If we move to, say, random Erdős-Rényi graphs with edge probability $1/2$, I would expect the eigenvalues to be distinct with high probability, though as far as I know proving this is still an open problem (EDIT: This was recently proved by Tao and Vu -- their preprint is here and is described in this blog post of Tao's).
For more on this issue, albeit for random preferential attachment trees instead of graphs, check out the recent paper Spectra of Large Random Trees by Bhamidi, Evans, and Sen.
Given a graph $G$, you can make a graph $H$ where the vertices of $H$ are the edges of $G$, and two vertices in $H$ are joined by an edge if and only if the corresponding edges in $G$ meet at a vertex of $G$. Then you can apply your favorite measure of centrality to vertices of $H$, and you'll actually be applying it to edges in $G$.
I don't know how good a measure of centrality you get this way, but I would think it would be worth a look.
Best Answer
One way to have high degree but low betweenness is if almost all of your friends know each other. This is because whenever you are between two other nodes, the shortest path must use two of your friends who are not friends with each other.
One way to have low degree but high betweenness is if your friends each have high degree, and know different people to each other.