There is such arrangement, for some large $k$. In the text below, I did not try to optimize $k$.
1. First, we reduce the problem to the following uncolored version:
There is an arrangement $\mathcal A$ of finitely many squares such that each $a\in\mathcal A$ overlaps with four pairwise disjoint squares in $\mathcal A$.
Indeed, given such arangement $\mathcal A$, for each square $a\in \mathcal A$ we introduce a color $C_a$, find four pairwise disjoint squares overlapping with $a$, and put into an arrangement $\mathcal B$ their copies having color $C_a$. Then $\mathcal B$ satisfies the requirements: each square in it is a copy of some $a\in\mathcal A$, and hence the color $C_a$ fits for it.
2. Now we construct a required $\mathcal A$. We introduce the colors to distinguish the squares; those colors have a few in common with the colors in the initial problem.
On each step of the process, we say that a square is good if it satisfies the required property, and bad otherwise. The degree of a square is the maximal number of its pairwise disjoint overlappers.
Start with a $10\times 10$ grid of just-non-overlapping red squares, and put into any interior vertex of the grid a blue square. All squares are good, except for the boundary red ones (see left figure).
A convenend property is that for any point $x$ in the upper-eft quarter of the figure (within the grid), we may take a large square with $x$ as an either upper-left, or upper-right, or lower-left corner; this square will be automatically good, as it overlaps with many red squares. We attach such squares (not shown in the figure) to each side of the grid. Now, still boundary red squares are bad, but each of them already has degree 3. [In fact, this is not needed, so it is made just for visibility reasons.]
Now take any two adjacent boundary red squares in the upper-left quarter; the right figure shows them magnified. We do the following chain of operations:
$\bullet$ add a green square; to two its dark-green vertices attach large non-overlapping squares, as described above. The red squares become good, the green one has degree 3 (due to the upper red square and two large oens);
$\bullet$ add a magenta square, and attach two large squares to its dark-magenta vertices --- its degree is 3 (due to the green square and two large ones), the green one becomes good;
$\bullet$ add a yellow square, attach three large squares. All squares in this small area are good (the yellow --- due to the upper red square and three large ones).
Applying such operations to every pair of adjacent boundary red squares (or even splitting them in pairs), we reach the goal.
(The construction is superfluous, but, as I told, I did not try to optimize.)