An example of a non-amenable exact group without  free subgroups. A countable discrete group $\Gamma$ is said to be exact if it admits an amenable action on some compact space. 
So clearly amenable groups are exact, but large familes of non-amenable groups are as well.
For many of the families that I know of (ex. linear groups, hyperbolic groups) that are exact, they also satisfy the von Neumann conjecture (i.e. that if they are non-amenable then they have subgroup isomorphic to a free group.)
So my questions is:
Are there examples of exact groups that are non-amenable and do not contain free subgroups?
 A: Owen, I'm a bit late to the party, but I think the answer to your question is ``no", to the best of my knowledge. To phrase it properly, I believe it is not known whether any of the known counterexamples to von Neumann's conjecture is exact. 
Jon, one has to be careful with limits hyperbolic groups, for example Gromov's random groups which are not exact are such limits (they are lacunary hyperbolic, in the sense of Olshanskii, Osin and Sapir). 
A: It has been some years since this question was posted, but maybe (if you haven't seen it yet) you'll enjoy reading the new geometric solution for the von Neumann problem.
http://www.math.cornell.edu/~justin/Ftp/vN_fp.pdf
A: I did not check the details, but most probably Gromov's random groups can be made torsion as well (and thus will not contain $F_2$). Just impose the relations $u^{n_u}$ on the steps with even numbers and do Gromov's construction of embedding the next graph of an expander on steps with odd numbers. Here $u$ runs over all words of the free group (more precisely, $u_k$ is the smallest length word that has infinite order in the group number $k-1$, and $n_{u_{k}}>>1$, odd, is chosen after the word $u_k$ is determined). Thus there are non-exact groups without free subgroups. If one does only the even steps of this construction, then the resulting group will be torsion and non-amenable (it will have the previous group as its factor) and most probably exact, although I am not as sure about it as about the non-exact example above. So the answer to the original question is most probably "yes". The way to prove exactness can be through finiteness of asymptotic dimension.  
