What is the minimum sampling rate for QPSK or what is the bandwidth of QPSK signal?
As best as I can tell from the terminology, QPSK a.k.a. 4-PSK or 4-QAM can convey twice the number of bits of information as BPSK in the same bandwidth.

Given a one-sided bandwidth of $B$ and assuming sufficient S/N ratio so that no bits are corrupted, BPSK can transfer bits at a rate of $2B$.  That means, given the same one-sided bandwidth, QPSK can transfer bits at a rate of $4B$.

But the sample rate must be at least $2B$.  If the receiver is perfectly synchronized to the transmitter, there would be one sample per bit with no bit error, if the transmitter sends a bipolar $\operatorname{sinc}(\cdot)$ for each bit.

But for QPSK, each sample is a complex value with a real part and imaginary part, each part containing one bit sample if this is 4-PSK.

So if you use quadrature demodulation to bump the IF or RF down to baseband $i(t)$ and $q(t)$, and sample those signals, you will need one complex-valued sample per pair of bits.