I believe theThe answer is no. Consider $\phi_0(x)=x^4+x^3$ on the field $F$ of size $q=2^7$. Then $\text{max}_a\lvert\phi_a(F)\rvert=83<2q/3$.
(Initially, 83 was 79, a miscalculation as pointed out by Boris Bukh.)
Actually, I believeif $\gamma>5/8$, then if $F$ is a field of order $q=2^r$ with $r$ odd and big enough, then $\text{max}_a\lvert\phi_a(F)\rvert\le\gamma q$.
Proof: Let $F$ be the field of order $q=2^r$, where $r$ is odd. It is easy to see that for each $\gamma>1-e^{-1}$$\lvert\phi_0(F)\rvert=q/2$. (EulerUpon setting $e=2.71\dots$$x=u+uv$, $y=uv$, the equation $\phi_0(x)=\phi_0(y)$ is equivalent to $u^3(v^2 + v + u + 1)$.) there
Thus the case $a=0$ is a mapfine. If $\phi_0$ on$a\ne 0$, and if $t$ is a suitable finite fieldtranscendental, then the Galois group of $F$$\phi_a(X)-t=X^4+X^3+aX-t$ over $\bar F(t)$ is the symmetric group $S_4$. Here $\bar F$ is the algebraic closure of even order $q$ such$F$.
Given that the Galois group is as claimed, an old result by Birch and Swinnerton-Dyer shows that $\text{max}_a\lvert\phi_a(F)\rvert\le\gamma q$$\lvert\phi_a(F)\rvert=(1-1/2!+1/3!-1/4!)q+O(\sqrt{q})$, where $O$ depends only on the degree, which is fixed here anyway. From $1-1/2!+1/3!-1/4!=5/8<2/3$ the claim follows.
(Initially So it remains to verify the Galois group: Using the Berlekamp discriminant, 83 was 79one can compute that $Gal(\phi_a(X)-t)$ contain odd permutations whenever $a\ne0$. Furthermore, a miscalculation as pointed outan easy computation shows that $\phi_a(X)$ is polynomially indecomposable over $\bar F$, so by Boris BukhLüroth the Galois group is primitive.) Well, degree $4$, primitive and not contained in $A_4$ implies $S_4$.
Comment(Reference: Tomorrow I hope to add some more detailsBirch, B. J.; Swinnerton-Dyer, H. P. F.: Note on a problem of Chowla. Acta Arith. 5 1959 417–423 (1959))