Is there ever a case where a comparison (equals()) between two floating point values would return false if you compare them as DOUBLE
Would I ever get an incorrect result if I promote 2 floats to double and do a 64bit comparison rather than a 32bit comparison?
No.
If you start with two floats, which could be float variables (float x = foo();) or float constants (1.234234234f) then you can compare them directly, of course. If you convert them to double and then compare them then the results will be identical.
This works because double is a super-set of float. That is, every value that can be stored in a float can be stored in a double. The range of the exponent and mantissa are both increased. There are billions of values that can be stored in a double but not in a float, but there are zero values that can be stored in a float but not a double.
As discussed in my float comparison article it can be tricky to do a meaningful comparison between float or double values, because rounding errors may have crept in. But, converting both numbers from float to double doesn't not change this. All of the mentions of epsilons (which are often but not always needed) are completely orthogonal to the question.
On the other hand, comparing a float to a double is madness. 1.1 (a double) is not equal to 1.1f (a float) because 1.1 cannot be exactly represented in either.