How this current mirror act as a replacement to resistor Rc?
The PNP transistor Q6 acts as an active load resistor and as a current mirror. Since current source Q6 has nearly infinite impedance, we will be able to have very large voltage gain Rc/r'e in which Q6 is the Rc.


However, intuitively speaking, I don't get how this current source will generate large gain.


For example, if v1(noninverting input) is at its negative peak, current passing through diode Q5 will decrease and this decrease in current will be mirrored to Q6. At the same time, Q2 collector current is increasing. How will that generate a negative peak AC voltage at Q2?

As the amplifier output is very high impedance, it's better to think of it as a current output amplifier, with the output voltage determined by the load it's driving. In normal use, the load will maintain the output voltage such that neither Q2 nor Q6 is saturated.
As Q6 current decreases and Q2 current increases, both of those effects mean that the output source current decreases, to the point of becoming an increasing sink current.

I mean, the Q6 has only a voltage drop of Vce at that point,

Q6 voltage drop is not Vce, the voltage is substantially determined by the load.

and I believe that won't be a sufficient drop to decrease the Vout to its negative peak. Just how does this current source act as a replacement to a normal resistor RC?

With a resistor in circuit to source the current, Q2 is working into the parallel impedance of the load and RC. Using Q6 as a current mirror, not only is the current gain doubled, but also the high output impedance of Q6 means that they are working into the load resistance only.
The voltage gain of the amplifier is given by its current gain times the load impedance.