As stated, your question is equivalent to the existence of a large exotic 4-ball (a smooth $D^4$ which cannot be smoothly embedded into $\mathbb{R}^4_{std}$).
The existence of a flip would give rise to an exotic $S^4$, by gluing the $D^4$ at infinity using the flip diffeomorphism. Removing a (small) standard ball from this $S^4$ gives a large exotic $D^4$, since it contains a smooth submanifold $K'$ which cannot embed in $\mathbb{R}^4_{std}$.
Conversely, if you had a large exotic $D^4$, then you could adjoin a collar neighborhood of $S^3\times \mathbb{R}$ to get an exotic $\mathbb{R}^4$ which has a standard end (diffeomorphic to $S^3\times \mathbb{R}$), and therefore admits a flip.
Although I'm not an expert, I'm certain that the existence of a large exotic 4-ball is open (otherwise, the 4D smooth Schoenflies conjecture would imply the 4D smooth Poincare conjecture).
I realize that this does not answer the spirit of your question, which is whether there is a large exotic $\mathbb{R}^4$ which does not have a standard product end and which admits a flip.
I wrote a little expository piece about this and related matters in the Newsletter of the European Mathematical Society:
http://www.ems-ph.org/journals/newsletter/pdf/2010-03-75.pdf
The classical topology of $X:=T^\ast L$ can be taken to include a little more than its diffeomorphism type: there's also an almost complex structure $J$ on $X$, canonical up to homotopy. The pair $(X,J)$ knows the Pontryagin classes of $L$, because $$c_{2k}(TX,J)|_{L}=c_{2k}(TL\otimes\mathbb{C})=(-1)^k p_k(TL),$$
so $(-1)^k p_k(TL)$ pulls back to $c_{2k}(L)$ under projection $X\to L$. However, even with this embellishment, the smooth topology of $X$ doesn't determine $L$.
Faithfulness conjecture: the exact symplectomorphism type of the cotangent bundle $(X,\omega=d\lambda_L)$ of a compact manifold $L$ determines $L$.
An exact symplectomorphism $T^\ast L \to T^\ast L'$ is a diffeomorphism $f$ such that $f^*\lambda_{L'}-\lambda_L= dh$ for $h$ a compactly supported function. The conjecture (but not the name) is standard.
Attempts to use symplectic invariants of $X$ to distinguish smooth structures on $L$ have so far been a complete failure. For example, the symplectic cohomology ring $SH^*(X)$ is isomorphic to loopspace homology $H_{-*}(\mathcal{L}L; w)$ (the coefficients are the local system $w$ of $\mathbb{Z}$-modules determined by $w_2$), with the string product. This invariant is determined by the homotopy type of $L$.
"Arnol'd's conjecture" (scare quotes because Arnol'd really made a much more circumspect conjecture). Any exact Lagrangian embedding $\Lambda \to X$ (with $\Lambda$ compact) is exact-isotopic to the embedding of the zero-section.
This would immediately imply the faithfulness conjecture.
There has been progress towards Arnol'd's conjecture of three kinds:
(1) It's true for $L=S^2$ (Hind, http://arxiv.org/abs/math/0311092).
(2) The work of several authors (Fukaya-Seidel-Smith http://arxiv.org/abs/0705.3450, Nadler http://arxiv.org/abs/math/0612399, Abouzaid http://arxiv.org/abs/1005.0358, and Kragh's work in progress) cumulatively shows that the projection from an exact Lagrangian to the zero-section is a homotopy equivalence. This is good evidence for the truth of the conjecture, but for the application to faithfulness one might as well make homotopy-equivalence an assumption.
(3) As Andy mentioned, Abouzaid http://arxiv.org/abs/0812.4781 has shown that a homotopy $(4n+1)$-sphere $S$, such that $T^*S$ contains an exact embedded Lagrangian $S^{4n+1}$, bounds a parallelizable manifold. This is proved by a stunning analysis of the geometry of a space of pseudo-holomorphic discs.
The existence of exact Lagrangian immersions is governed by homotopy theory (there is an h-principle which finds such an immersion given suitable homotopical data). Just as the subtleties of 4-manifold topology can be located at the impossibility of removing double points of immersed surfaces (the failure of the Whitney trick), so the subtlety of the symplectic structure of cotangent bundles comes down to the question of removability of double points of Lagrangian immersions.
Best Answer
This is problem 4.79(A) of Kirby's 1995 problem list, contributed by Gompf. It is my impression that it is still open.
There is some small progress: Remark 7.2 in this article observes that their constructions imply that a specific countable set of examples of exotic $\Bbb R^4$s cannot possibly cover a closed manifold. This is not a huge reduction, as only countably many exotic $\Bbb R^4$s could possibly be covers of the countable set of closed smooth 4-manifolds, anyway. But at least the examples are somewhat explicit.
I couldn't find any other references to this problem in the literature, but that doesn't mean there aren't any.
UPDATE: I emailed Bob Gompf; he is not aware of any recent progress.