The answer would be positive if there would be no condition on rationality of the coefficients. With this condition the answer is negative. Indeed, assume first that $f_5$ is not constant. Then $f_5(t) = 0$ for some $t \in \mathbb{Q}$. Substituting it into the equation you get $\sum f_i^2(t) = 0$, hence each $f_i(t) = 0$ as we are over rationals. It follows that all $f_i$ are proportional to $f_5$. On the other hand, if $f_5$ is constant then by looking at the leading coefficients you see that the sum of their squares is zero, so each coefficient is zero, and so all $f_i$ are constant too.

Note that the same argument works for arbitrary number of squares and for any degree of polynomials.

Over $\mathbb{C}$ it is easy to give an example. Take $f_1 = x$, $f_2 = \sqrt{-1} x$, $f_3 = 0$, $f_4 = f_5 = 1$.

The answer would be positive if there would be no condition on rationality of the coefficients. With this condition the answer is negative. Indeed, assume first that $f_5$ is not constant. Then $f_5(t) = 0$ for some $t \in \mathbb{Q}$. Substituting it into the equation you get $\sum f_i^2(t) = 0$, hence each $f_i(t) = 0$ as we are over rationals. It follows that all $f_i$ are proportional to $f_5$. On the other hand, if $f_5$ is constant then by looking at the leading coefficients you see that the sum of their squares is zero, so each coefficient is zero, and so all $f_i$ are constant too.

Note that the same argument works for arbitrary number of squares and for any degree of polynomials.

Over $\mathbb{C}$ it is easy to give an example. Take $f_1 = x$, $f_2 = \sqrt{-1} x$, $f_3 = 0$, $f_4 = f_5 = 1$.