Let $\mathfrak{p} \subseteq \mathbb{Z}[X_0,\dots,X_n]$ be a homogeneous prime ideal generated by some homogeneous, irreducible polynomials $P_1,\dots,P_r$. Pick a prime number $p$ that does not divide any coefficient of the $P_i$'s. Is it true that the homogeneous ideal $\mathfrak{p} + (p)$ is still prime?
Some background: I am trying to prove that, given a projective, irreducible curve defined over $\mathbb{Q}$, there are only finitely many primes $p$ such that the "corresponding" curve over $\mathbb{F}_p$ is not irreducible. This should be true, but I am not sure whether my argument is correct. Certainly one needs the assumption that $p$ does not divide the coefficients (or something similar), because for instance $XY^2-pZ^3$ is irreducible but $(XY^2-pZ^3,p)=(XY^2,p)$ is not prime. We may assume that $\mathfrak{p}$ does not contain constants, if this helps.
Your first question is answered negatively in the comments.
As for your background question, it doesn't work either. Basically, we can mimic the proof that $X^4+1$ is reducible mod $p$ for any prime $p$, but with the homoegenous polynomial $X^4+Y^4$, which is an irreducible element of $\Bbb Z[X,Y]$ (see here). We prove that for any prime $p$, the reduction $X^4+Y^4 \in \Bbb F_p[X,Y]$ is reducible.
If $-1$ is a square in $\Bbb F_p$ (which includes the case $p=2$), say $a^2=-1$, then we have $$X^4+Y^4=X^4-a^2Y^4=(X^2+aY^2)(X^2-aY^2).$$
If $p$ is odd and $2$ is a square in $\Bbb F_p$, say $2=b^2$, then we have $$X^4+Y^4=(X^2+Y^2)^2-(bXY)^2=(X^2+bXY+Y^2)(X^2-bXY+Y^2). $$
If $p$ is odd and neither $-1$ nor $2$ is a square, then their product $-2$ is a square, say $-2=c^2$. (since $\Bbb F_p^\times$ is a cyclic group of even order). Then we have $$ X^4+Y^4=(X^2-Y^2)^2-(cXY)^2=(X^2-cXY-Y^2)(X^2+cXY-Y^2).$$