Are exotic affine spaces motivic/whatever equivalent to affine space? This question is inspired by this MO question; in turn by this MO; in turn by these MO, MO.
An exotic affine space is an affine variety $V$ whose $\mathbb{C}$-points are diffeomorphic to $\mathbb{R}^{2n}$ yet $V$ is not algebraically isomorphic to $\mathbb{A}^n$.
Will Sawin showed in his answer MO that  the number of $F_q$ points is the same  for fake affine spaces as for non-fake (for generic $q$).
Question: Are exotic affine spaces equivalent to affine space in Grothendieck ring of varietes ? Or may be there are  some other simple geometric equivalence which is stronger than just point counting and weaker than isomorphism ?  
PS
Evgeny Shinder in comments to MO states that " Among nonsingular projective varieties, a fake projective plane or odd-dimensional quadrics have the same point count as projective planes. In the case of a fake projective plane its class in the Grothendieck ring is not L^2 + L + 1 (and in fact not congruent to 1 (mod. L) because it's not stably rational). The class of a quadric is same as [P^n] (using projection from the point). "
 A: Here is an argument showing that if $V$ is a smooth complex surface with trivial integral homology groups (note that exotic $\mathbf{A}^2$ do not exist, as explained in the comments), then $[V] = \mathbb{L}^2$ in the Grothendieck ring of varieties. We use the notation $\mathbb{L} = [\mathbb{A}^1]$.
It follows from this paper: https://projecteuclid.org/euclid.jmsj/1230128845
(which I learnt about from this MO post Topologically contractible algebraic varieties) that $V$ is a rational surface.
Let $X$ be a non-singular compactification of $V$ such that the divisor $D = X \setminus V$ has simple normal crossings with smooth components $C_1, \dots, C_t$. 
Since $X$ is a non-singular projective rational surface, we have $[X] = 1 + k\mathbb{L} + \mathbb{L}^2$. We have $[D] = [C_1] + \dots + [C_t] + r$, where $r \in \mathbb{Z}$ ($r$ depends on the intersection graph of these curves). 
Altogether $[V] = 1 + k\mathbb{L} + \mathbb{L}^2 - [C_1] - \dots - [C_t] - r$. Applying Hodge realization to this equality, using the fact that $V$ has a trivial Hodge structure, we deduce that all curves $C_i$ are rational so that $[C_i] = [\mathbb{P}^1] = 1 + \mathbb{L}$. Thus $V$ is a polynomial in $\mathbb{L}$ and the only possibility is $[V] = \mathbb{L}^2$.
P.S. If $V$ an exotic affine space of dimension one, then $V$ is isomorphic to $\mathbb{A}^1$, because it has to be a genus zero curve. If $V$ is an exotic affine space of dimension $3$ or higher, rationality of $V$ is an open question.
