SiGe/GaAs RF MMIC as photodiode preamp
I'm afraid the simple answer is no. Even assuming 100% QE, one photon produces one electron. What sort of current does this imply?

Well, one ampere is one coulomb per second, and $$k_e = 1.6 \times 10^{-19} C$$ Applying Ohm's Law is not straightforward here, since detecting a single photon is clearly a short-term event. But let's say that, since the amplifier bandwidth is 100 MHz, the arrival of an electron can be considered a current pulse with a width of 100 nsec. Then $$V = iR = \frac{\Delta Q}{\Delta t}\times R = \frac{1.6 \times 10^{-19}}{10^{-8}}\times 50 = 8\times 10^{-10} 
\text{ volts}$$ While this is obviously pretty small, you need to compare it to the amplifier noise, and assuming the amp is Johnson noise limited with an effective temperature of 300 C, $$V_n = \sqrt{4k_{B}RT\times BW} = \sqrt{4\times1.37\times10^{-23}\times300\times50\times10^{8}} = \sqrt{82.2\times 10^{-12}} = 9\times 10^{-6}\text{ volts}$$

In this calculation, the expected signal is about 10,000 times less than the rms noise voltage, and detection would be, shall we say, challenging.

This ignores shot noise, which makes things worse, but this seems perfectly reasonable in view of the already-unfavorable situation.