Let $K$ be a $(p,q)$ torus knot on the torus $T_1$. Via the map \begin{equation*} H=\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \end{equation*} $K$ becomes a $(q,p)$ torus knot on $T_2=H(T_1)$. If we fill in both these tori to become solid tori $D_1$ and $D_2$, then \begin{equation*} S^3 = D_1\cup_H D_2 \end{equation*}
This seems to give a Seifert fibering of $S^3$, and the comments to this question suggest that's true. But wouldn't the base orbifold be a a 2-sphere with two conical points of order $p$ and $q$, which is a "bad" orbifold? I also know that if we remove a tubular neighborhood of $K$ in $S^3$, then we do get a Seifert fibering of the knot complement.
I'd like to figure out where my understanding is breaking down. Is this an actual Seifert fibering of $S^3$ or not?
The base space of a Seifert fibered 3-manifold can indeed be a bad 2-orbifold, so your intuition is not breaking down. See this math overflow answer for reference including a discussion which incorporates bad base orbifolds. So yes, your description of the base orbifold associated to a $(p,q)$ torus knot is correct.