Leading off with the following.
If $g$ is a primitive element of a finite field then multiplication by $g$ takes a non-zero square to a non-zero non-square and vice versa.
So if we can represent $0$ as a sum of two non-squares, then multiplication by $g$ shows that $0$ can also be written as a sum of two squares:
$$
0=a^2+b^2.
$$
This implies that
$-b^2=a^2$ and a fortiori that $-1=(a/b)^2$ is a square. This is known to be the case in $\Bbb{F}_q$ if and only if $q\equiv1\pmod4$.
Consequently:
There exists arbitrarily large finite fields such that the element $0$ cannot be written as a sum of two non-squares of that field. More precisely, this happens in the field $\Bbb{F}_q$ whenever $q\equiv-1\pmod4$.
A more interesting result is that any non-zero element $z$ of a finite field $\Bbb{F}_q$, $q$ and odd number $>5$, can be written as a sum of two non-squares. This can be seen as follows.
Assume first that $z$ is a square. Then $g^{-1}z$ is a non-square. By the well- known result we can write it as a sum of two squares
$$
g^{-1}z=x^2+y^2.
$$
Because $g^{-1}z$ is a non-square, we can deduce that $x$ and $y$ must both be non-zero. This means that the elements $gx^2,gy^2$ are both non-squares, and
$$
z=gx^2+gy^2
$$
is a presentation of the required type.
If $z=ga^2$ is a non-square then we need the result (see e.g. Ireland and Rosen) that the equation
$$
x^2+y^2=1\qquad(*)
$$
has $q-\eta(-1)$ solutions (here $\eta$ is the unique multiplicative character of order two, so equal to the Legendre symbol in the case of a prime field). The equation $(*)$ is equivalent to
$$
a^2x^2+a^2y^2=a^2,
$$
so the equation $x^2+y^2=a^2$, too, has $q-\eta(-1)\ge q-1$ solutions. At most $4$ of those solutions have either $x=0$ or $y=0$. So if $q>5$, then we are guaranteed the existence of elements $x\neq0\neq y$ such that
$g^{-1}z=a^2=x^2+y^2$. Again, it follows that
$$
z=gx^2+gy^2
$$
is a presentation of $z$ as a sum of two non-square.
The OP noted themself that in the fields of $3$ or $5$ elements there are too few non-squares. For example in $\Bbb{F}_5$ the only non-squares are $2$ and $3$, and we cannot write either of those as sums of two non-squares.
For the case of the prime fields the elegant solution by Mikhail Ivanov is surely better than this argument.
Best Answer
No, this is not the case. The number of such possible sums $\le N$ for large $N$ is far less than $N$.
The number of powers of $2$ (or of $3$ or $5$) up to $N$ is $O(\log N)$. Hence the number of products $2^a 3^b 5^c$ up to $N$ is $O((\log N)^3)$. The number of sums of pairs of such products is $O((\log N)^6)$, which is $o(N)$.