Effect of different electron withdrawing/transition state stabilising groups in nucleophillic aromatic substitution
Your reasoning is correct.  The relative reaction rates would probably be quite similar if we didn't have the methyl groups on both sides of the substituent X.  

In order for the substituent to stabilize the transition state through resonance, the substituent must be coplanar with the aromatic ring.  This isn't a problem for the linear $\ce{CN}$ group, but the $\ce{NO2}$ group with the two oxygens protruding out will not be able to remain planar with the aromatic ring.  The two adjacent methyl groups create steric interference with the planar nitro group; this steric interaction is minimized when the nitro group is twisted out of the aromatic plane.  When the nitro group is twisted out of plane it cannot resonance stabilize the transition state.  Hence the nitro compound reacts more slowly than the cyano compound. 

While the underlying reaction is different, the thinking behind this question is identical to that found in this question from yesterday.