Can I use a chi-squared test to compare two empirical distributions?
Signs point to "yes." R's chisq.test allows two vectors x and y. It says:
cases with missing values are removed, the objects are coerced to factors, and the contingency table is computed from these. Then Pearson's chi-squared test is performed of the null hypothesis that the joint distribution of the cell counts in a 2-dimensional contingency table is the product of the row and column marginals.
See also: same question about KS test.