No, consider the map $F_3 \to F_3$ given by $$ x_1 \mapsto x_1, x_2 \mapsto x_2[x_3, x_1], x_3 \mapsto x_3 [x_1, x_2[x_3, x_1]]. $$ It is an IA automorphism: it clearly induces the identity map on abelianization. Further, it is an isomorphism because the set $\{ x_1, x_2[x_3, x_1], x_3 [x_1, x_2[x_3, x_1]] \}$ generates and $F_3$ is Hopfian.
Further, if you kill $x_2$, you get the map $$ x_1 \mapsto x_1, x_3 \mapsto x_3 [x_1, [x_3, x_1]], $$ which is clearly not an isomorphism (the word $x_3 [x_1, [x_3, x_1]] = x_3 x_1^{-1} (x_3^{-1} x_1^{-1} x_3 x_1) x_1 (x_1^{-1} x_3^{-1} x_1 x_3)$ in reduced form begins and ends with $x_3$).