A nice graph theory lemma for showing homotopy equivalence that builds on the "clique starring" already discussed is stated as Lemma 3.2 of Alexander Engström's paper arXiv:math/0508148. I'll rephrase in terms of the language the question was asked in (clique complexes rather than independence complexes).

**Lemma:** If $v$ and $w$ are vertices of a graph $G$ with $N[v] \subseteq N[w]$, then $C(G)$ is homotopy equivalent to $C(G \setminus v)$.

(Here $N[v]$ is the closed neighborhood of $v$, i.e., $v$ and all its neighbors.)

The proof technique is that of elementary collapses. See
http://en.wikipedia.org/wiki/Collapse_(topology).

Even if the lemma of Engström doesn't give you what you need, the broader technique of collapsing can be quite useful. Collapses are the "engine" of discrete Morse theory, for example.

It's not too hard to write a computer program (or some are available) that does automatic collapsing, and you could generate a large number of examples this way. If you can collapse a complex to a point, then the complex is contractible. The converse is not true, but the program might at least give you a smaller complex to examine by hand.

On the other hand, as Benjamin Steinberg points out, the barycentric subdivision of any simplicial complex is flag, so classifying contractible flag complexes should be as hard as classifying contractible simplicial (and more generally CW) complexes.