Not necessarily as written in the generality you want. Suppose that we are in $\mathbb R^1$, there is no linear constraint, and $Q(y)$ is some positive function of $y$. Then $v(y)\equiv 0$, which is convex. Now add the condition $x<-1$. Then $v(y)=Q(y)$ and that can be absolutely anything.
On the other hand, you, probably, know that $v(y)$ is convex for some particular reason coming from certain properties of $Q,E,d$ (I'm not yet sure the ones you listed will suffice in higher dimensions, so you'd better tell the whole story of how you prove the convexity and what you use there) and we can certainly discuss whether in that particular situation restricting $x$ to a fixed convex polyhedron (not necessarily bounded) will preserve convexity or even the convexity proof.
Edit: With arbitrary $A$ and $b$ you are still in a bad shape. Indeed, let $n=2$, $x_1,x_2$ be the partition of $x$ into the vectors of length $2$, and suppose that $Ax\le b$ is equivalent to $x_1=x_2$. Let $c=0$. Then you have no choice but to take $x_1=x_2=\frac y{y_1+y_2}$. Now let $F_1=I_2, F_2=0$. Then $$ v(y)=y_1\frac{y_1^2+y_2^2}{(y_1+y_2)^2} $$ which on the interval $y_1=t, y_2=1-t$ ($0<t<1$) is $t[t^2+(1-t)^2]=2t^3-2t^2+t$, so it is obviously not convex near $0$. Thus, some restrictions on $A$ and $b$ should be added before we can hope for a positive answer. What can you offer?