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I don't think this will hold

EDIT: Rephrased the answer in light of Koose's comment.

Claim. Any ideal satisfying the required property has to be a complete intersection.

Proof. Let $\mathfrak p\subset R$ be a prime ideal. Assume that the minimal number of generators for $\mathfrak p$ is $r$ and let $a_1,\dots,a_r\in\mathfrak p$ be a set of generators. Let $I_t=(a_1,\dots,a_t)$ and one has the sequence of ideals:

$$0\subsetneq I_1 \subsetneq \dots \subsetneq I_t\subsetneq I_{t+1}\subsetneq \dots\subsetneq I_r=\mathfrak p$$

The containments cannot be equalities, because that would make the corresponding $a_i$ unneeded to generate $\mathfrak p$. If all the $I_t$'s are prime, then $\mathfrak p$ has height at least $r$, but it cannot be more than that, so the claim is proven. $\square$

Example. Take an irreducible projective variety (say the twisted cubic curve) that is not a complete intersection (in projective space). Then the ideal of the affine cone over this in the affine cone over projective space will be a prime, but if you take away one of the generators, that will not be.

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I don't think this will hold. Take an irreducible projective variety (say the twisted cubic curve) that is not a complete intersection. Then the ideal of the affine cone over this in the affine cone over projective space will be a prime, but if you take away one of the generators, that will not be.