Electrical – Is it sufficient to disconnect the negative terminal of a voltage source

There are a couple of issues that you need to be aware of.

First of all, you need to make sure that they share the blocking voltage equally. Just like with diodes, you need to put high-value resistors in parallel with their drain and source terminals such that the current through the resistors is on the order of 10× the reverse leakage current of the transistors (or diodes). This will equalize the voltage drops to within 10% or so. Also, the speed of the individual devices might vary a bit, so it's also a good idea to put capacitors (typically 1 to 10 nF in line-frequency applications, less at higher frequencies) in parallel with them as well, to tame the transient behavior.

The second big issue is that the gate drive to each transistor needs to be relative to that transistor's source terminal; they can't share a common connection. Depending on the application, you might use transformers, optoisolators or specialized high-side driver ICs to allow each of the gate drive signals to float relative to all the others.

The body terminal is there whether we like it or not because of the construction of a MOSFET:

The gate voltage modifies the channel width. But the gate potential is measured relative to the substrate potential, and the channel is formed in the substrate material. Therefore the MOSFET behavior depends strongly on the substrate (body) potential.

If we did not make a contact to the body, it would be free to float, and we could not control the behavior of the transistor. We connect to the body to take control of this behavior. Normally we tie the body to either a most-positive or most-negative potential, like you say, because that gives the lowest |V_gs| relative to the appropriate supply and prevents forward biasing the source-body junction. If we did not contact the body, it would still be there, but we would not have control over its effect.

(MOSFET image by Wikimedia user CyrilB licensed under CC-BY-SA)