Why 'the electric current is “shorted out” by the conductor' in Stutzman-Thiele Antenna Theory and Design?
To answer your question directly: No, that is not true.

A perfect electric conductor(PEC) can sustain a current on the surface, in fact, it is very good at doing this since it is perfectly conducting! The reason you can ignore \$J_s\$ in this case is because a current density can not radiate on a PEC. Think of it in terms of image theory, the tangential electric field produced by \$J_s\$ will be canceled out by its image producing an opposite electric field, see Figure 1. So for our purposes of solving for the radiated fields \$E\$ and \$H\$, \$J_s\$ can be ignored (ie. \$J_s\$ = 0). However, you can also see that a magnetic current density \$M_s\$ producing a tangential magnetic field would essentially be doubled on a PEC rather than canceled! From electromagnetic uniqueness concepts we know that we only need \$M_s\$ (or\$J_s\$) to solve for both the magnetic and electric fields. Therefore we use our a priori knolwedge of the tangential fields over the surface \$S\$ (remember we didnt know the sources within, but we did have to know the tangential fields on \$S\$) and solve for \$M_s\$ using:
$$
M_s = E\times \hat{n}
$$

For a simple geometry such as a flat PEC as previously discussed we get simply:
$$
M_s = 2E\times \hat{n}
$$

See Figure 2.

This also fits very well with a simple case of an open waveguide propagating a TE mode since you have known tangential electric fields and presumably a large conducting flange around the mouth of the waveguide. You can carry out this same procedure, see Figure 3.



Figure 1 - Image Theory PEC



Figure 2 - Equivalent Magnetic Current on PEC



Figure 3 - Equivalence Applied to Waveguide