Radar Pulse Compression Gain (PCR)
The answer is yes but one has to specify $B_n$ properly to avoide possible confusions. In case if one uses a pulse compression, the bandwidth through which the receiver collects the noise will normally be $B_n = \beta_c$. Then, the "new" signal-to-noise ratio should be written as:

$SNR = \dfrac{P_TG_TG_R\lambda^2\sigma{P_g}}{(4\pi)^3R^4(kT_{sys}\beta_c)} = \dfrac{P_TG_TG_R\lambda^2\sigma{\tau_p\beta_c}}{(4\pi)^3R^4(kT_{sys}\beta_c)} = \dfrac{P_TG_TG_R\lambda^2\sigma{\tau_p}}{(4\pi)^3R^4(kT_{sys})}$

As you see, the signal-to-noise ratio remained the same as for the pulse without modulation (for which $B_n = 1/\tau_p$). 

So, the pulse compression helps to increase the range resolution but the signal-to-noise ratio remains the same as for the unmodulated pulse. This holds true if you do not change $\tau_p$ and the peak power $P_T$ when you modulate your pulse.

The confusion (and as you mentioned "conflicting opinions") comes from the fact that it is quite often forgotten that if you increase the bandwidth of your pulse then you also increase the bandwidth through which the receiver collects the noise since radar's receiver is matched to the pulse and therefore has a bandwidth $\beta_c$. 

Your last equation is correct. But since you put $F_n$ - the noise figure- then it should be specified that this is the signal-to-noise raito at the output of the receiver that has a noise figure $F_n$.