Let $p$ be an odd prime and $F={\mathbb Z}/p{\mathbb Z}$. With $a,b,c\in F$, let $Q(x,y)=ax^2+bxy+cy^2$ where $p\nmid b^2-4ac$. I wish to prove that the number of solutions $(x,y)\in F^2$ of $$ax^2+bxy+cy^2=u$$ is the same for all units $u\in F$.
For example, when $p=5$, the equation $x^2-2xy+3y^2=u$ has one solution when $u=0$ and 6 solutions for each of $u=1,2,3,4$.
Let $F$ be a finite field, such as $\mathbf Z/(p)$ but it could be a more general finite field (including of characteristic $2$, and not necessarily $\mathbf Z/(2)$, so I won't make a change of linear variables using division by $2$ as in another answer). If both $a = 0$ and $c = 0$ then the equation becomes $bxy = u$, so necessarily $b \not= 0$ since $b^2 - 4ac = b^2$ and you said $b^2 - 4ac \not= 0$. A solution $(x,y)$ has both coordinates not equal to $0$, with $y = u/(bx)$. Thus the solution set for each $u \in F^\times$ is $\{(x,u/(bx)) : x \in F^\times\}$, whose size is $|F| - 1$.
If $a \not= 0$ or $c \not= 0$ then by symmetry (swapping the roles of $x$ and $y$ in the equation) we can suppose $a \not= 0$. Then solving $ax^2 + bxy + cy^2 = u$ is the same as solving $x^2 + (b/a)xy + (c/a)y^2 = u/a$, and as $u$ runs over $F^\times$ so does $u/a$. Thus we can suppose $a = 1$: we want to count solutions of $x^2 + bxy + cy^2 = u$ in $F^2$. Set $R = F[t]/(t^2 + bt + c)$, a finite ring. The norm map ${\rm N}_{R/F} \colon R \to F$ is multiplicative, and using the basis $\{1,t\}$ we can realize your expression for $Q(x,y)$ as a norm value: for $x, y \in F$, $$ {\rm N}_{R/F}(x + yt) = \det\begin{pmatrix}x&-cy\\y&x-by\end{pmatrix} = x^2 - bxy + cy^2 = (-x)^2 + b(-x)y + cy^2. $$ Therefore the equation $x^2 + bxy + cy^2 = u$ is the same as ${\rm N}_{R/F}(-x+yt) = u$ for $x, y \in F$. On units the norm map ${\rm N}_{R/F} \colon R^\times \to F^\times$ is a group homomorphism, so as with all homomorphisms between finite groups, all values are taken on an equal number of times. Thus it remains to show the norm map ${\rm N}_{R/F} \colon R^\times \to F^\times$ is surjective.
Case 1: $t^2 + bt + c$ is irreducible in $F[t]$. Then $R$ is a field, so $R^\times$ is cyclic and the norm map $R^\times \to F^\times$ on the nonzero elements of finite fields is onto (if $|F| = q$ then $|R| = q^2$ and a generator of $R^\times$ is mapped to a generator of $F^\times$).
Case 2: $t^2 + bt + c$ is reducible in $F[t]$. Write it as $(t-r)(t-s)$. Since $b^2 - 4c = (r-s)^2$, from $b^2 - 4c \not= 0$ we have $r \not= s$. Then $R \cong F[t]/(t-r) \times F[t]/(t-s) \cong F \times F$, and in the basis $\{(1,0),(0,1)\}$, the norm mapping has the formula ${\rm N}_{R/F}(x,y) = xy$, which maps $R^\times = F^\times \times F^\times$ onto $F^\times$.
You definitely want $b^2 - 4ac \not= 0$, since if $b^2 - 4ac = 0$ your desired conclusion can fail. As a simple example, consider $Q(x,y) = x^2 + 2xy + y^2 = (x+y)^2$. Then the equation $Q(x,y) = u$ has solutions in $F$ if $u$ is a nonzero square in $F$ and it has no solutions in $F$ if $u$ is a nonzero nonsquare in $F$. When $F = \mathbf Z/(p)$ for odd $p$ (or more generally $F$ is a finite field of odd characteristic), half of $F^\times$ is squares and half is nonsquares, so the equation $(x+y)^2 = u$ for $u \in F^\times$ has solutions for half the values of $u$.
The following works for any finite field $F$. Consider the projective curve $$ C : ax^2 + bxy + cy^2 = uz^2. $$ Your conditions $b^2-4ac\in F^*$ and $u\in F^*$ imply that $C$ is non-singular, since it has degree $2$, the genus of $C$ is $0$, and elementary (or non-elementary) methods show that $C(F)\ne\emptyset$, so $C\cong\mathbb P^1_F$, where the isomorphism is defined over $F$. Hence $$ \#C(F) = \#\mathbb P^1(F) = |F|+1. $$ So the question becomes whether the point(s) at infinity, i.e., the points with $z=0$, are in $C(F)$. This is clearly independent of $u$, and it depends on whether $b^2-4ac$ is a square in $F$. Hence $$ \#\bigl\{ (x,y)\in F : ax^2+bxy+cy^2=u\bigr\} = |F| - \left(\frac{b^2-4ac}{F}\right), $$ where $$ \left(\frac{A}{F}\right)=\begin{cases}1&\text{if $A\in{F^*}^2$,}\\ -1&\text{if $A\notin{F^*}^2$.}\\ \end{cases} $$
Inspired by @JoeSilverman's answer, I've updated this answer to handle any odd-characteristic field $F$, and to remove a few ugly case distinctions from the original argument; but I do assume that the characteristic is not $2$, as @KConrad points out.
Put $\theta = \dfrac{b + \sqrt{b^2 - 4a c}}2$, where we have chosen a square root of $b^2 - 4a c$ that does not equal $-b$ (either one, if $4a c \ne 0$).
We have that $a x^2 + b x y + c y^2 = a x^2 + (\theta + a c\theta^{-1})x y + c y^2$ equals $(a x + \theta y)(x + c\theta^{-1}y)$. If $\theta$ belongs to $F$, then the change of variables $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & \theta \\ 1 & c\theta^{-1} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$ is non-singular (with determinant $a c\theta^{-1} - \theta = (a c - \theta^2)\theta^{-1} = \sqrt{b^2 - 4a c}$), so we are considering solutions $(x', y')$ of $x'y' = u$, which form an $F^\times$-torsor. (In particular, if $F$ is finite, then there are $\lvert F^\times\rvert$ of them.)
If, on the other hand, $\theta$ does not belong to $F$, then $b^2 - 4a c$ is not a square in $F$, so $a$ is non-$0$. Then $a x^2 + bx y + cy^2$ equals $a^{-1}(a x + \theta y)(a x + a c\theta^{-1} y) = a^{-1}N_{F[\theta]/F}(a x + \theta y)$, so parameterising solutions of $a x^2 + b x y + c y^2 = u$ is the same as parameterising solutions of $N_{F[\theta]/F}(a x + \theta y) = u a$. Since $(x, y) \mapsto a x + \theta y$ is an isomorphism of $F$-vector spaces (not of $F$-algebras) $F \oplus F \to F[\theta]$, we are parameterising the fibre of the norm map over $u a$. In general, of course, fibres are either empty or torsors for $\ker N_{F[\theta]/F}$; and, if $F$ if finite, so that the norm map is surjective, then all fibres are torsors, hence have size $\lvert F[\theta]^\times\rvert/\lvert F^\times\rvert$.
