Phase Comparator and VCO in PLL
The output of the phase comparator is the same : the output of the "loop filter" is not. 

To achieve zero phase error, you need infinite gain at zero frequency; i.e. a component of the loop filter is an integrator.

An easier way to think of the loop filter, in this case, is as a PI controller, with the P (Proportional) term providing fast tracking of phase variations, and the I (Integral) term eliminating phase error.

The P term alone would leave some residual phase error, which multiplied by the "P gain", would provide the DC voltage required to drive the VCO.

EDIT to preserve OP's quote from comments...


  At steady state condition the phase error is 0, but that value enters
  an integrator. An integrator saturates at DC but in this case its
  input value is not simply DC, it is 0. Therefore, its output value
  will be the initial value (vout = v(t0) + integral(0 × dt) = v(t0),
  where t0 is the instant in which lock is reached). So, in cases f1 and
  f2, v(t0) is different because of their different story, so a
  different value will drive the VCO.


Yes.