Interesting question. I would say from an energy standpoint, it almost certainly it has no effect.
Of course, the extreme case is Io, one of the Galilean moons whose heat source comes from the gravitational tidal stretching as it orbits very closely to the planet Jupiter. The heat that sustains the core of the Earth, however, is left over from its formation and also comes from radioactive decay of heavy elements.
The differential potential energy (and hence the tidal force) over the planet Io due to Jupiter, which is approximately 1300 times more massive than the Earth, is much larger than that of the Earth due to the moon. The relationship between force and differential potential energy is:
$$ F = - \nabla U $$
At a given location on the potential energy curve, the strength of the force is determined by its steepness (derivative) at that same location. Below is a quick plot I've generated for the Earth-Moon system, where the vertical red line represents the average Earth-Moon distance over a one-year period. As you can see it doesn't appear to be very 'steep', though keep the scales of the x- and y-axes in mind.
Admittedly this isn't that exciting of a plot. But, for comparison one could be made for the Jupiter-Io system, and numerical derivatives could be taken for both to calculate the magnitude of the tidal force in each situation.
To answer the question:
If the difference in the gravitational potential energy of object A on B over the scale of B is comparable to the self-gravitational energy of object B, then tidal forces will become important. This self-gravitational energy is the amount required to completely pull apart all massive particles infinitely far away. Formally, this limit is called the Roche Limit.