Yes, this is possible, you can even find a $C^\infty$-smooth map with this property.
Let $C\subset R^n$ be a closed subset. A $C^k$-smooth function $F: C\to R^m$ is
a function such that for every $a\in C$ there exists a neighborhood $U$ of $a$ in $R^n$ and a $C^k$-smooth extension of $F|_{C\cap U}$ to $U$.
Just to settle the terminology, I will be considering the $n$-dimensional (hyper)cube $Q$ given by the inequalities
$$
|x_i|\le 1, i=1,...,n.
$$
Then the $2^n$ regions $P_i, i=1,...,2^n$, of $Q$ appearing in your question are the connected components of the complement $Q-D$, where
$$
D= \{(x_1,...,x_n): x_1\cdots x_n=0\}.
$$
(Each $P_i$ is a unit cube in $R^n$ missing part of its boundary.)
In the answer to your previous question you were given a $C^\infty$-function $f: Q\to R^m$ such that the images of the cubes $P_i\subset Q$ are pairwise disjoint. I will construct a $C^\infty$-smooth extension of $f$ to a surjective function $F: R^n\to R^m$.
Consider the $n$-dimensional orthant
$$
O_n= \{x\in R^n: x_i\ge 2, i=1,...,n\}.
$$
Lemma. For every $m\le n$, there exists a surjective $C^\infty$-smooth map $g: O_n\to R^m$.
Proof. Consider the function $\phi: R\to R$,
$$
\phi(t)= t^{k+1}\sin(t).
$$
This function is easily seen to be $C^\infty$-smooth and surjective. Moreover, its restriction of $\phi$ to each interval $[T,\infty)$ is also surjective.
Now, define
$$
g(x_1,...,x_n)= (\phi(x_1),...,\phi(x_m)).
$$
This function $R^n\to R^m$ is also clearly $C^\infty$-smooth and satisfies $g(O_n)=R^m$. qed
Now, for the set $C=Q\cup O_n$ we define a function $h: C\to R^m$,
$$
h(x)= \begin{cases}
f(x), \quad x\in Q\\
g(x), \quad x\in O_n.
\end{cases}
$$
By the construction, this function is surjective, $C^\infty$-smooth and the images of cubes $P_i$ are pairwise disjoint.
The last ingredient we need is Whitney extension theorem which is a smooth analogue of Tietze Extension Theorem. I will formulate only a weak version of Whitney's theorem which is much easier to prove (the full Whitney's theorem also prescribes values of partial derivatives).
Theorem. Suppose that $A\subset R^n$ is a closed subset and $f: R^n\to R^m$ is a $C^k$-smooth function, $1\le k\le \infty$. Then $f$ admits a $C^k$-smooth extension $F: R^n\to R^m$.
Remark. Whitney's theorem is usually stated only for $R$-valued functions but applying it to each component $f_i, i=1,...,m$ of the function $f: A\to R^m$, one obtains the theorem stated above.
Lastly, applying Whitney's theorem to the function $h$ defined above, we obtain a surjective $C^\infty$-smooth map $F: R^n\to R^m$ extending the map $h$ and, therefore, satisfying the property that the images of the cubes $P_i$ are pairwise disjoint. qed
Here is what I do not know:
Question. Is there an open surjective smooth (or even continuous) map $F: R^n\to R^m$ such that the images of the cubes $P_i$ are pairwise disjoint?
