I think the answer is yes when $n=2$.

Let $R=\mathbb{Q}[x,y]$; since $R$ is a two-dimensional the only non-trivial case is for height-1 primes, and since $R$ is a UFD we can suppose that the prime ideal $P$ is generated by an irreducible polynomial $f(x,y)$. We distingush two cases.

**Case 1:** $f$ is linear. Since $P\cap V=(0)$, there are $\alpha,\beta,\gamma\in\mathbb{Q}$, with $\gamma\neq 0$ (since $P\cap V=(0)$), such that $f(x,y)=\alpha x+\beta y+\gamma$. By applying a linear substitution (which does not change $V$), we can suppose that $f(x,y)=y+\gamma$. Let $g\in\mathbb{Q}[z]$ be an irreducible polynomial of degree $>1$, and define $M:=(g(x),y+\gamma)R$.

**Case 2:** $f$ is not linear. Without loss of generality, suppose that the $x$-degree of $f$ is $>1$. By Hilbert's irreducibility theorem, there are infinitely many $y_0\in\mathbb{Q}$ such that $g(x):=f(x,y_0)$ is irreducible; in particular, we can choose $y_0$ different from $0$ and such that the degree of $g$ is $>1$. Define $M:=(g(x),y-y_0)R$, and note that $P\subseteq M$ (write every $y^k$ of $f(x,y)$ as $(y_0+(y-y_0))^k$: the $y_0^k$ goes to build $g$, while all the rest is a multiple of $y-y_0$).

In both cases, we have $M:=(g(x),y-y_0)R$ for some $y_0\neq 0$; I claim that $M\cap V=(0)$. Clearly $x\notin M$ (otherwise $g(0)\in M$, against the fact that $g$ is irreducible and not linear). Suppose $M$ contains $\alpha x+y$: then, $\alpha x+y-(y-y_0)=\alpha x+y_0\in M$. Hence, $M$ should contain both $g(x)$ and $\alpha x+y_0$, which is impossible since (again) $g$ is irreducible of degree $>1$.

Therefore, $M$ is your maximal ideal.

For $n>2$, I think a similar proof should work, with $M$ being in the form $(f_1(x_1),f_2(x_2),\ldots,f_{n-1}(x_{n-1}),f_n(x_n))$, with each $f_i$ irreducible and at most one of them linear; however, you probably need some better control on degrees of the generators of $P$ (for example, a generator of $P$ may be linear in two variables but in the other ones).