Should isometries be linear?

I suppose you meant that $T$ is affine rather than merely linear. So, suppose $T(0)=0$ and hence $Q(T(u))=Q(T(u)-T(0))=Q(u-0)=Q(u)$. We want to know if $T$ is linear or not.

As you said that $Q$ is nondegenerate, I also suppose that $\operatorname{char} F\ne2$ and $Q$ is induced by a symmetric bilinear form $b(x,y)=\frac12\left(Q(x+y)-Q(x)-Q(y)\right)$. It follows that $b(T(x),T(y))=b(x,y)$ for all $x,y\in V$.

Since every symmetric bilinear form on a finite dimensional vector space over a field of characteristic $\ne2$ is diagonalisable, we may further assume that $V=F^n$ and $b(x,y)=x^\top Dy$ for some diagonal matrix $D$. As $Q$ is nondegenerate, $D$ is invertible.

Let $\{e_1,\ldots,e_n\}$ be the standard basis of $F^n$ and let $\mathbf{T}=(T(e_1),\ldots,T(e_n))\in M_n(F)$. By assumption, for any $x\in V$, we have $T(e_i)^\top DT(x) = b(T(e_i),T(x)) = b(e_i,x) = e_i^\top Dx$ for each $i\in\{1,2,\ldots,n\}$. Therefore \begin{align*} &\mathbf{T}^\top DT(x)=I_nDx, \textrm{ for all } x\in V,\tag{1}\\ &\mathbf{T}^\top D\mathbf{T}=D.\tag{2} \end{align*} The equality $(2)$ implies that $\mathbf{T}$ is invertible. Hence $(1)$ implies that $T(x)=D^{-1}\mathbf{T}^{-\top}Dx$, i.e. $T$ is linear.