By De Gua's Theorem we know the areas of the three right angle faces $a,b,c$ and the base $d$ are related by an extension of the Pythagorean Theorem to trirectangular tetrahedrons:
$$a^2+b^2+c^2=d^2$$
Is it possible to have all the face areas be odd integers? A quick check of all integers less than 1000 does not turn up any examples. In fact, there seem to be only two cases by examining solutions. Either (1) a triangular tetrahedron must have two right angle faces with even area and one right angle face and base with odd area or (2) all faces have even area.
No, the face areas, i.e., $a$, $b$, $c$ and $d$, can't all be odd integers in
$$a^2 + b^2 + c^2 = d^2 \tag{1}\label{eq1A}$$
This is because $n^2 \equiv 1 \pmod 4$ for all odd integers $n$. Thus, the left side would be congruent to $3$ modulo $4$, but the right side would be congruent to $1$ modulo $4$.