Do I need to duplicate the 4th "substrate" connection when building CMOS gates out of discrete transistors?
The short answer is that you don't need 4 terminal FETs to build CMOS logic.

Some Background:

In a simple CMOS process, (P-type Wafer, N-Wells) the substrate contact is directly connected to the conductive wafer. This means that the body terminal of all NFETs are basically shorted together. A similar effect happens with the PFETs, although it isn't as absolute. They aren't shorted together to improve performance, but because it is cheaper and easier to manufacture.

This brings up a question: If we had to tie the body terminal of all NFET devices together, what voltage would we like it to be? For NFETs, the body-source and body-drain connections normally look like reverse-biased diodes. In order to maintain these diodes reverse-biased, the body voltage must be less than \$V_D \mbox{ or } V_S + 0.6\mbox{ Volts}\$. Typically this is done by tying the substrate/body to the most negative voltage present in the system. In digital systems, this is usually ground \$V_{SS}\$. The body terminal of PFETs is typically tied to the most positive voltage, or \$V_{DD}\$ for similar reasons.

For 3-terminal FETs, where the source and body have been internally shorted, the internal diodes will never be forward-biased if the source is always at a lower voltage than the drain. If you are stuck with 4-terminal transistors building discrete gates, it will work with the bodies connected to \$V_{SS}\$ and \$V_{DD}\$, and it will also work with the body shorted to the source.