I think that this is wrong for $n = 3$:

Let $C \subset {\mathbb P}^3_{\mathbb R}$ be a generic cubic surface not containing $(0:0:0:1)$ and take $$Q \colon x^2 + y^2 + z^2 = \varepsilon w^2$$ with $\varepsilon > 0$ sufficiently small (the coordinates are $x,y,z,w$). Then $C({\mathbb R}) \cap Q({\mathbb R})$ is empty, but for most choices of $C$ (and $\varepsilon$) $V = C \cap Q$ will be integral over $\mathbb C$.

For larger $n$, the same example works (the real points that show up will be singular).

If the signature of $Q$ is $(2,2)$ instead of $(3,1)$, then the statement is true: $Q$ is isomorphic over $\mathbb R$ to ${\mathbb P}^1 \times {\mathbb P}^1$ and $V$ gets mapped to a curve of type $(3,3)$, which will always have lots of real points. So perhaps one needs an additional assumption on the signature of $Q$.

I think that this is wrong for $n = 3$:

Let $C \subset {\mathbb P}^3_{\mathbb R}$ be a generic cubic surface not containing $(0:0:0:1)$ and take $$Q \colon x^2 + y^2 + z^2 = \varepsilon w^2$$ with $\varepsilon > 0$ sufficiently small (the coordinates are $x,y,z,w$). Then $C({\mathbb R}) \cap Q({\mathbb R})$ is empty, but for most choices of $C$ (and $\varepsilon$) $V = C \cap Q$ will be integral over $\mathbb C$.

For larger $n$, the same kind of example works.

If the signature of $Q$ is $(2,2)$ instead of $(3,1)$, then the statement is true: $Q$ is isomorphic over $\mathbb R$ to ${\mathbb P}^1 \times {\mathbb P}^1$ and $V$ gets mapped to a curve of type $(3,3)$, which will always have lots of real points. So perhaps one needs an additional assumption on the signature of $Q$.