Is the surface of a heavy sphere bigger than $4 \pi r^2$ due to general relativity?
Yes. The radial distance you measure does not jive as you move outwards. In other words if you move out by 100km the surface area will not be 4π(R + 100km)2, but a little less.
There are two 'r' here. One is the distance covered on the trip from the centre to the surface (say 'r'), the other is determined by measuring the area of the planet and using the 4πR2 rule. These will not be equal as they would in flat space. r will be bigger than R.
Or is this exactly backwards from what you are thinking? I guess it depends on which r you are using in your question.
Short answer is yes (at least according to GR): despite the fact that your reasoning is not rigorous, you came to the right conclusion.
It can be seen mathematically. Take one of the idealized solutions of GR (the most simple one is the Schwarzschild metric) and evaluate the integral over the sphere with radial coordinate $r$. Then integrate the distance from $0$ to $r$ - the radius of the sphere (note that it is not equal to $r$ since the radial coordinate is not physical - it is just a quantity we introduce). Compare these results and it will lead you to the same conclusion.