Do different $R$-forms for the same simple $FG$-module have the same vertex?

Question

Let $G$ be a finite group. Let $R$ be a discrete valuation ring with residue field $k$, where $k$ has positive characteristic $p$. Let $F$ be the field of fractions of $R$.

Let $V$ be a simple $FG$-module. We say that an $RG$-module $U$ is an $R$-form for $V$ if $F\otimes_R U\cong V$.

Must every $R$-form for the same simple $FG$-module $V$ have the same vertices?

What I know:

1. All $R$-forms of $V$ must lie in the same $p$-block, so this gives a common upper bound for their vertices.
2. Let $P$ be a Sylow $p$-subgroup of $G$. If $V_P$ is free as an $FP$-module, then $V$ is the only member of its $p$-block, and any $R$-form of $V$ must be projective.

Answer 1

No, this is not true. Consider the group $G=\Sigma_5$ with $p=2$, and suppose that $R$ has enough roots of unity. Then there is a (unique up to isomorphism) $6$ dimensional simple $FG$-module. It is induced from a $3$ dimensional simple for $A_5$, so there is an $R$-form whose vertex is a Sylow subgroup of $A_5$. However, there is also an $R$-form whose reduction mod two is a direct sum of two one dimensional summands and a four dimensional summand, so this $R$-form has to have vertex a Sylow $2$-subgroup of $G$. For this, see Lemma I.18.2 of Feit's book, "The representation theory of finite groups."