The answer is no. Consider $\phi_0(x)=x^4+x^3$ on the field $F$ of size $q=2^7$. Then $\text{max}_a\lvert\phi_a(F)\rvert=83<2q/3$.
Actually, I believe that for each if $\gamma>1-e^{-1}$ (Euler \gamma>5/8$, then if$e=2.71\dots$) there F$ is a map $\phi_0$ on a suitable finite field $F$ of even order $q$ such that 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$is odd. It is easy to see that$\lvert\phi_0(F)\rvert=q/2$. (Initially, 83 was 79Upon setting$x=u+uv$,$y=uv$, the equation$\phi_0(x)=\phi_0(y)$is equivalent to$u^3(v^2 + v + u + 1)$.) Thus the case$a=0$is fine. If$a\ne 0$, and if$t$is a miscalculation transcendental, then the Galois group of$\phi_a(X)-t=X^4+X^3+aX-t$over$\bar F(t)$is the symmetric group$S_4$. Here$\bar F$is the algebraic closure of$F$. Given that the Galois group is as pointed out claimed, an old result by Boris Bukh.) Comment: Tomorrow I hope Birch and Swinnerton-Dyer shows that$\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. So it remains to add some more detailsverify the Galois group: Using the Berlekamp discriminant, one can compute that$Gal(\phi_a(X)-t)$contain odd permutations whenever$a\ne0$. Furthermore, an easy computation shows that$\phi_a(X)$is polynomially indecomposable over$\bar F$, so by Lüroth the Galois group is primitive. Well, degree$4$, primitive and not contained in$A_4$implies$S_4$. (Reference: Birch, B. J.; Swinnerton-Dyer, H. P. F.: Note on a problem of Chowla. Acta Arith. 5 1959 417–423 (1959)) 2 added 127 characters in body I believe the answer is no. Consider$\phi_0(x)=x^4+x^3$on the field$F$of size$q=2^7$. Then$\text{max}_a\lvert\phi_a(F)\rvert=79\text{max}_a\lvert\phi_a(F)\rvert=83<2q/3$. Actually, I believe that for each$\gamma>1-e^{-1}$(Euler$e=2.71\dots$) there is a map$\phi_0$on a suitable finite field$F$of even order$q$such that$\text{max}_a\lvert\phi_a(F)\rvert\le\gamma q$. (Initially, 83 was 79, a miscalculation as pointed out by Boris Bukh.) Comment: Tomorrow I hope to add some more details 1 I believe the answer is no. Consider$\phi_0(x)=x^4+x^3$on the field$F$of size$q=2^7$. Then$\text{max}_a\lvert\phi_a(F)\rvert=79<2q/3$. Actually, I believe that for each$\gamma>1-e^{-1}$(Euler$e=2.71\dots$) there is a map$\phi_0$on a suitable finite field$F$of even order$q$such that$\text{max}_a\lvert\phi_a(F)\rvert\le\gamma q\$.