Alright, putting together all the comments it appears to still be true when $q = 2$, depending on how you like your square-free polynomials.

First, to quote Sonia's comment, If $P$ is not sq.free, then for each sq.free polynomial $Q$ of degree $n$, we have a non-zero $P−Q$, of degree $\leq n−1$, $2^{n−1}$ in all. The number of sq.free polynomials of degree $\leq n−1$ is $2^{n−1}$. So there must be nonempty intersection. – Sonia Balagopalan

Now, suppose $P$ is square-free and has degree $n$. For any square-free polynomial of lesser degree, $Q$, $P-Q$ is a polynomial of degree $n$. There are $2 + 2 + 2^4 + \ldots + 2^{n-2} = 2^{n-1}$ choices for such $Q$, but there are only $2^{n-1}$ non-square-free polynomials of degree $n$ and we certainly did not land on $0$, so at least one of these must be another square-free polynomial of degree $n$.

So any polynomial of degree $n \geq 3$ can be written as the sum of two square free polynomials of degrees $n$ and $d$ where $d < n$. What about when $n = 2$? The only polynomials to consider are

$P_1 = t^2,$

$P_2 = t^2 + 1,$

$P_3 = t^2 + t$ and

$P_4 = t^2 + t + 1$.

If you want to add the restriction that our two square free polynomials MUST have degree at most 2, as Sonia claims, $P_3$ and $P_4$ are counterexamples. Otherwise we may write

$P_1 = (t^2 + t) + (t)$,

$P_2 = (t^2 + t) + (t + 1)$,

$P_3 = (t^2 + t^7 + 1) + (t + t^7 + 1)$ and

$P_3 = (t^2 + t^7 + t) + (t^7 + 1)$.

(Note: There is nothing too special in picking the $7$.)

Upon further thought, you can do what you want with the degree 1 polynomials. A similar trick using a larger degree polynomials will work, or you can use those as counterexamples to a restricted degree version if you like. Although trivial, you may also want to express $0$ and $1$ as sums, which can certainly be done as $z$ and $z+1$ are both square-free.