I believe the answer is yes.  Let $X\neq \emptyset$ and let $P,Q,R\in {\rm Part}(X)$.

First, note that if $|P|<|Q|$, then $P<_{\rm eff}Q$, since $|P-Q|\leq |P|<|Q|=|Q-P|$.

Now we are ready for the proof.  Assume $P<_{\rm eff}Q<_{\rm eff}R$.  Without loss of generality, we may remove $P\cap Q\cap R$ from all the sets, and reduce to the case that $P\cap Q\cap R=\emptyset$.  By our previous paragraph, we also have $|P|\leq |Q|\leq |R|$.  If inequality holds anywhere, then we have $P<_{\rm eff}R$ and we are done.  Assume the contrary case holds, and so all three sets have equal cardinality.

From the fact that $|P-Q|<|Q-P|$ and $|P|=|Q|$, we see that $|P\cap Q|=|P|=|Q|$.  Now because $(P\cap Q)\cap R=\emptyset$, we have $|Q-R|\geq |P\cap Q|=|Q|=|R|\geq |R-Q|$, contradicting the fact that $Q<_{\rm eff}R$.  Thus, this contrary case cannot actually happen after all.