I believe one can deduce an improvement of the form 
$$I(P,L) \leq |P| |L|^{1/2-\epsilon}$$
from the symmetric case $|P|=|L|$ by using the following argument of [Pudlak][1] (see Corollary 2.5, there): Assume that $|P| > |L|$ and select a random subset $P'$ of $P$ of size $L$. The expected number of incidences will be $I(P,L) \frac{|L|}{|P|}$. Let $P'$ denote a set of points at or above the expectation. Now applying the symmetric result to the symmetric incidence problem with $P'$ and $L$, should give 
$$I(P,L) \frac{|L|}{|P|} \leq I(P',L) \leq |L|^{3/2-\epsilon'}. $$

In the case of large sets, there is also a [result of Le Anh Vinh][2], which states that:

$$I(P,L) \leq q^{-1} |P| |L| + q^{1/2} |P|^{1/2} |L|^{1/2}. $$ 

Here, $q$ is the order of the finite field. Note, however, that this is worse than trivial when $|P|\times|L|$ is smaller than $q$.


  [1]: http://link.springer.com/chapter/10.1007/3-540-33700-8_11
  [2]: http://arxiv.org/pdf/0711.4427v1.pdf