Let $\mathrm{Even}(n)$ be the number of colorings of the $1\times n$ board in which the number of red and of green squares is each even. Let $\mathrm{Odd}(n)$ be the number of colorings of the $1\times n$ board in which the number of red and green squares is each odd.
For $n\geq 2$, color the initial $1\times (n-2)$ board any which way; there are $4^{n-2}$ ways of doing it. Now, if the number of red tiles and the number of green tiles are both even (which happens in $\mathrm{Even}(n-2)$ of the colorings), then we can complete the coloring by either coloring both remaining tiles green, both red, or each of them either blue or orange. This gives six different colorings with the same "base" coloring on the initial $1\times (n-2)$ board.
If the number of red tiles and the number of green tiles are both odd (which happens in $\mathrm{Odd}(n-2)$ of the colorings), then we must color the remaining two tiles one red and one green, in some order. That's two different colorings with the same "base" coloring.
If one of the number of red tiles and of green tiles is even and the other is odd (which occurs in $4^{n-2} - (\mathrm{Odd}(n-2)+\mathrm{Even}(n-2))$ of the colorings), then one of the two remaining tiles must be the deficient color, and the other tile can be either blue or orange. That gives 4 possible completions (select which tile is either blue or orange, select the color).
A similar computation holds for $\mathrm{Odd}(n)$, since we are trying to preserve parity.
So we have the recursion
$$\begin{align*}
\mathrm{Even}(n) &= 6\mathrm{Even}(n-2) + 2\mathrm{Odd}(n-2) + 4\bigl(4^{n-2}-\mathrm{Odd}(n-2)-\mathrm{Even}(n-2)\bigr)\\
&= 4^{n-1} + 2\mathrm{Even}(n-2) - 2\mathrm{Odd}(n-2)\\
&= 4^{n-1} + 2\Bigl(\mathrm{Even}(n-2) - \mathrm{Odd}(n-2)\Bigr).\\
\mathrm{Odd}(n) &= 6\mathrm{Odd}(n-2) + 2\mathrm{Even}(n-2) + 4\bigl(4^{n-2}-\mathrm{Odd}(n-2) - \mathrm{Even}(n-2)\bigr)\\
&= 4^{n-1} + 2\Bigl(\mathrm{Odd}(n-2) - \mathrm{Even}(n-2)\Bigr).
\end{align*}$$
Thus, we have:
$$\mathrm{Even}(n) - \mathrm{Odd}(n) = 4\Bigl(\mathrm{Even}(n-2)-\mathrm{Odd}(n-2)\Bigr).$$
We also have $\mathrm{Odd}(0) = 0$, $\mathrm{Even}(0)=1$; and $\mathrm{Odd}(1)=0$ and $\mathrm{Even}(1)=2$.
From this, it should be easy to get a formula for $\mathrm{Even}(k) - \mathrm{Odd}(k)$, and from there a formula for $\mathrm{Even}(n)$ for all $n\geq 1$.
Best Answer
The best way is using the principle of inclusion exclusion.
There are $24^n$ ways to color the board so no columns is monochrome; $3\times 3\times 3-3$ ways to color each column, except for the three ways where that column is monochrome.
From these, we must subtract the colorings where no column is monochrome, but some row is monochrome. There are three choices for which row is monochrome, three choices for its color, then $3\times 3-1=8$ ways to fill each column. Therefore, it seems like the answer is $24^n-3\times 3\times 8^n$.
However, colorings with two monochrome rows have been double subtracted. These must be added back in. There are $\binom32=3$ ways to choose to the two rows.
If the two rows are the same color, there are $3$ ways to choose the common color, and $2^n$ ways to fill in the rest of the board.
If the rows are different colors, there are $3\cdot 2$ ways to color them, then $3^n$ ways to color the last cell in each row.
We are currently at $24^n-3\times 3\times 8^n+\binom32\times\Big(3\times 2^n+ 3\times 2\times 3^n\Big)$. We must now consider colorings where all three rows are monochrome. These were triple subtracted in the first step, then added back in three times in the previous step, so these need to be subtracted out again. There are only $24$ such colorings. The final answer is therefore $$ 24^n-9\times 8^n+18\times 3^n+9\times 2^n-24 $$