triode would conduct too much current in the face of a positive rail bounce, as its grid will see the positive pulse, while its cathode is being pulled away by the negative rail bounce. With roughly twice the grid-to-cathode voltage, its current increase will be roughly equal 2Gm.
     If we can move slightly back in this direction, then we can match the 1 / (Ra + rp) increase from the top triode. Moving back slightly is accomplished by retaining some small bypass capacitor (C2) to the negative rail. By varying the value of this capacitor, we can tune the ratio of power supply noise out of the output. What value should we start at for capacitor C2? If it had terminated into ground, C1 / mu would be a good starting point, where C1 is the capacitor that connects to the positive rail. But as this capacitor terminates into the negative power supply rail, C1 / 2mu is a better starting point. In other words, some tweaking will be required.
    One thought about your amplifier is that it could be differently arranged. Rather than use two feedback loops, use only one. (The White CF comprises its own feedback loop.) This could be accomplished by making the output a simple push-pull one fed from a Differential amplifier, with a twists. The big problem with a tube push-pull output stage is that one tube works as a Cathode Follower while the other works as a

Grounded Cathode amplifier. Two solutions present themselves: make the top tube function as a Grounded Cathode amplifier or make the bottom tube function as a Cathode Follower. I prefer the latter approach, as it lessens the need for global feedback.
    One way to make the bottom triode function as a Cathode Follower is to apply 100 percent degenerative feedback from its plate to its grid. Thus any perturbation at the output will be relayed un-attenuated to its grid and the triode will effectively have become a Cathode Follower in gain and output impedance. How do we apply 100 percent degenerative feedback? One easy way is to reference a high impedance gain stage to the output so that whatever signal is injected at the output is relayed back down to the bottom triode's grid.
      In the following schematic, the first triode of the Differential amplifier finds a MOSFET source at the end its plate resistor. If the shunting capacitor that bridges the output to this MOSFET's gate had gone to ground instead, the MOSFET stage would constitute a simple quasi-voltage regulator. But attaching as it does to the output, the MOSFET's source traces the output signal from the amplifier and this tracing moves in anti-phase to the signal present at the plate of the first triode: 100 percent feedback.

     Consequently, the gain of the triode is greatly reduced. This gain reduction performs two essential functions. The first is that it balances the drive voltages to both triodes. The top triode's grid sees a signal voltage only slightly higher than the output signal. For example, if working into a 300 ohm load results in a gain of 0.9 for the top triode, then when the grid sees a +10 volt pulse, the output swings +9 volts in response and the