Here is how the examples in TI 809-04 (the Army Corps of Engineers' "seismic design for buildings"

handle this question.
Say that your tall CMU load bearing wall gymnasium has a flexible roof diaphragm. In this case, the base shear in each direction would be determined (using the appropriate R value for the lateral system in that direction) based on the roof diaphragm's self weight and the weight of the perpendicular walls. Then, when designing the shear wall for in-plane forces, that wall's own inertial forces would be added to the distributed base shear forces. You add the inertial forces AFTER you've distributed the base shear because there's no way that the inertial forces in one end wall could suddenly jump out of the wall and be transmitted through the flexible diaphragm and into the other end wall.
Now, for a rigid diaphragm, there is the possibility that the inertial forces in a wall could be distributed to another parallel wall. In the case of a rigid diaphragm, the base shear is calculated based on the weight of the entire building including both parallel walls and perpendicular walls. (In this case, WHYUN, you are correct, the only difference between the base shears in either direction would be the R values.) This total base shear is then distributed based on the walls' relative rigidities.
So, in both cases, the inertial forces in the walls are being accounted for SOMEWHERE. In the inital post by wailuku, it appears this person was asking if the forces due to parallel walls can be omitted. Obviously, they must be accounted for somewhere. It is really just a matter of answering the question, "Can the inertial weight of a wall be distributed to other parallel walls, or must the wall be designed to resist it's own entire inertial weight?"
I hope this helps!