Relative size of the 2 tidal bulges
Yes, to some degree, but not for exactly the reason you're imagining.
Even in the idealized model you're using, the size of the tides is not controlled by the strength of the moon's gravitational pull at $A$ and $C$, but by the difference in the moon's gravitation between $A$ and $B$, or between $C$ and $B$.
This difference in gravitation between two neighboring points fall off proportionally to the inverse cube of distance rather than inverse square of the graviational field itself.
Since the moon is about 30 earth diameters away, we would expect the tidal effect to be about one part in 90 less on the far side of the earth.
In reality the "tidal bulge" model is very inaccurate for explaining how the oceans rise and fall. It is more accurate to consider the oceans as a very complex driven oscillator, where the water is slowly sloshing around. The periodic minor variations in gravity caused by the moon contribute energy to the sloshing, but the actual shape and timing of the sloshes is much more influenced by coastlines than than by the moon itself.
Indeed tides in the earth's oceans do have a component with a period of 25 hours (the approximate time from moonrise to moonrise) as well as the more well-known component with a period of 12½ hours. In some places (such as the Gulf of Mexico) the 12½-hour mode almost vanishes, such that even though it is driven at 90 times the strength globally, the amplitude of the 25-hour oscillation dominates locally. This leads to diurnal tides.
The strength of the gravitational force falls off like the square of the distance, $r^{-2}$.
Tides arise because the gravitational force on an extended object is stronger on the near side than it is on the far side. That difference goes like the rate of change of the force, so the tidal stretching goes like $r^{-3}$.
An asymmetry in the tidal force would take place if the tidal stretching on one side of your extended object were larger than the stretching on the other side. The size of this asymmetry would vary like the rate of change of the strength of the tidal force, or like $r^{-4}$.
So yes, there should be such an asymmetry. But if the magnitude of the tidal stretching is small compared to the gravitational force, the magnitude of the asymmetry will be small compared to the tides by kind of the same factor.