This is an interesting question.
Lets start with the assumption that if a bike (specifically the bike's center of gravity) is able to travel in a straighter line over uneven terrain, it will be more efficient and/or faster.
The pertinent question then becomes, how does sag allow the bike to travel in a straighter line?
Suppose that a unicyclist (or consider only half of a bike for simplicity) is riding on a suspended wheel with an effective spring rate that is linear, such that at rest for every 50 lbs of rider weight, the center of gravity (CoG) of the rider sags down by 1 inch. (k=50lbs/in). So a 150 lb (F = 150) rider will sag 3 inches (x=3 in) on the cycle while moving over perfectly flat terrain.
Now, for the sake of an example, say there is a 2 inch step down in the terrain. With no sag, as the riders wheel leaves the upper part of the step the horizontal component of velocity will remain unchanged, but in the vertical direction, the CoG will accelerate under gravity until the wheel contacts the lower step.
With sag, the rider CoG will still accelerate under gravity, but the wheel will also be accelerated away from the CoG and towards the ground under the (initially) 150 lb spring force. The wheel will then contact the ground before the un-sagged riders wheel would. Lets say that after 0.07 seconds the rider CoG has moved 1 inch down, and the wheel has been pushed another 1 inch down. At this point the suspension is preloaded by 100 lbs ([3 initial sag 1 unloading]*50 lbs/in = 100 lbs ). This 100 lbs of force pushes upwards on the riders CoG and slows the downward component of the riders acceleration and effectively straightens out the riders path.
We can also see that as soon as the wheel touches the ground in this example, the rider has 100 lbs force acting into the ground that can be used to steer or brake.