It seems to me that there is an obvious Whitehead-manifold inspired construction to try here to produce a counterexample. Let w : S^1 x D^2 -> S^1 x int D^2$w : S^1 \times D^2 \to S^1 \times \operatorname{int} D^2$ be the familiar Whitehead embedding, which is null-homotopic but not “null-isotopic”, for which the two inclusions S^1 x bdy D^2 \subsetinclusion S^1 x D^2 \setminus w(S^1 x int D^2)$S^1 \times \partial D^2 \hookrightarrow S^1 \times D^2 \setminus w(S^1 \times \operatorname{int} D^2)$ and w(S^1 x bdy D^2) \subsetinclusion S^1 x D^2 \setminus w(S^1 x int D^2)$w(S^1 \times \partial D^2) \hookrightarrow S^1 \times D^2 \setminus w(S^1 \times \operatorname{int} D^2)$ are each pi_1$\pi_1$-injective (google Whitehead manifold.) For i = 1$i = 1$ and 2 let w_i : T^2 x D^2$2$ let -> T^2 X D^2$w_i : T^2 \times D^2 \to T^2 \times D^2$ be the two obvious product embeddings engendered by w$w$, namely, let w_2 : = iden(S^1) x w : S^1 x S^1 x D^2 -> S^1 x S^1 x int D^2$w_2 : = \operatorname{iden}(S^1) \times w : S^1 \times S^1 \times D^2 \to S^1 \times S^1 \times \operatorname{int} D^2$, and similarly let w_1 = “w x iden(S^1)$w_1 =$ “$w \times \operatorname{iden}(S^1)$” (the quotes meaning that you do the appropriate factor permutations here to make it work). It looks as though the contractible direct-limit 4-manifold M^4$M^4$ gotten from the sequence w^1$w^1$, w^2$w^2$, w^1$w^1$, w^2$w^2$, … is not simply-connected at infinity (like the Whitehead 3-manifold). True?
