Electronic – Gain in Sallen-Key filter

One ting that jumps out is you have not selected a model for the diode - right click and select one from the list. Otherwise it may not simulate correctly.

I think what you are seeing may be to do with the rectifier not being able to sink current (only a small amount will flow through the 200k resistor), so the input signal will not look as it would unloaded, it will "charge" to it's peak. Then the larger the cap, the longer the fall time when the signal stops. Try putting a non-inverting buffer in between the rectifier and Sallen Key input. This should make both filters respond in the same way.

Let's look at a Sallen-Key low-pass filter:

One thing you will notice: the filter does not introduce any additional DC path to ground. C2 is connected to "ground", but since there is no DC path to it, it doesn't actually matter where it's connected, as long as it's a fixed voltage. We could just as well connect it to \$V_{CC}\$, or any other power rail. It doesn't matter, except for power-on transients.

How about a high-pass filter?

Here, we have a path to ground through R2, but R2 is 10kΩ. The point of a virtual ground IC is to provide a low impedance virtual ground, but here we need a 10kΩ ground. We don't need an IC for that, we just need a voltage divider made of two 20kΩ resistors. Sure, you could use a virtual ground IC and follow it with a 10kΩ resistor, but what's the point? A pair of 20kΩ resistors is a lot simpler.

Look at the Sallen-Key topology in general:

In this topology, there is always some impedance (\$Z_4\$) between the filter and ground. Since the point of a virtual ground IC is to make a low impedance ground, but we would never need that, the Sallen-Key "negates the requirement of a Virtual Ground". In other words, it isn't that you couldn't use a virtual ground IC: it's that you'd never need to use one.