Take any two closed connected homeomorphic smooth closed 4-manifolds that are not diffeomorphic. Then their products with $\mathbb R$ are diffeomorphic because the smooth structure on a 5-manifold is unique. It follows that the original manifolds are tangentially homotopy equivalent, i.e. there is a homotopy equivalence that pulls stable tangent bundles to each other. A priori this homotopy equivalence need not be homotopic to a homeomorphism but if one of your manifold is stably parallelizable, so is the other one, and then the homeomorphism has to preserve the stable tangent bundle because the pullback of a trivial bundle is trivial.

Take any two closed simply-connected homeomorphic smooth closed 4-manifolds that are not diffeomorphic. Then their products with $\mathbb R$ are diffeomorphic because the smooth structure on a such a product is unique. (Indeed, since PL/O is 6-connected, it is enough to show that the associated PL structure is unique, but the set of PL-structures on a PL-manifold $M$ of dimension $\ge 5$ is bijective to the set of homotopy classes of maps from $M$ to $TOP/PL$, and the latter space is $K(\mathbb Z_2, 3)$, so the set of PL structures on $M$ is bijective to $H^3(M,\mathbb Z_2)$, which vanishes by Poncare duality if $M$ is homotopy equivalent to a simply-connected $4$-manifold; in fact the argument shows that all we need is $H_1(M;\mathbb Z_2)=0$). 

It follows that the original closed simply-connected $4$-manifolds are tangentially homotopy equivalent, i.e. there is a homotopy equivalence that pulls stable tangent bundles to each other.