Let $G$ be a finite group and let $p$ be a prime dividing the order of $G$. Let $\chi$ be a $\mathbb{C}_p$-valued irreducible character of $G$ and let $e_{\chi} = |G|^{-1}\chi(1)\sum_{g \in G} \chi(g^{-1})g$ be the associated primitive central idempotent in $\mathbb{C}_p[G]$. Let $\mathbb{Q}_{p}(\chi)=\mathbb{Q}_p(\chi(g) : g \in G)$ be the character field. Let $H=\mathrm{Gal}(\mathbb {Q}_{p}(\chi)/\mathbb{Q}_p)$ and let $e=\sum_{h \in H} e_{\chi^h}$ ($H$ acts on characters in the usual way.) Then $e$ is a central primitive idempotent of $\mathbb{Q}_p[G]$. Let $v_p$ denote the usual $p$-adic valuation.

Claim: $v_p(|G|)=v_p(\chi(1))$ if and only if $e \in \mathbb{Z}_p[G]$ and $e\mathbb{Z}_p[G]$ is a maximal $\mathbb{Z}_p$-order.

If $v_p(|G|)=v_p(\chi(1))$ then it is clear that $e \in \mathbb{Z}_p[G]$. That $e\mathbb{Z}_p[G]$ is a maximal $\mathbb{Z}_p$-order follows from Jacobinski's formula for the central conductor of $\mathbb{Z}_p[G]$ in a maximal order (see Curtis-Reiner, Methods of representation theory, vol 1 section 27).

For the converse, I can prove the claim for $p$ odd again using Jacobinski's formula and some calculations of the different of the extension $\mathbb{Q}_p(\chi)/\mathbb{Q}_p$.

Question: can anyone provide a proof of counterexample for the missing part for $p=2$?

Here is a related claim that would prove the claim and make everything much simpler if true: If $e \in \mathbb{Z}_p[G]$ then $\mathbb{Q}_p(\chi)/\mathbb{Q}_p$ is unramified (i.e. $\mathbb{Q}_p(\chi) \subseteq \mathbb{Q}_p(\zeta_n)$ for some $n$ relatively prime to $p$).

Also, maybe I can drop the maximal order part of the claim altogether?

I have a reasonable knowledge of ordinary representation theory but have only really started to look at modular representation theory in the past few days. I know that this is related to "blocks of defect zero", but in the books I have looked at (Serre, Curtis-Reiner) it is assumed that the ground field is "sufficiently large", which doesn't really help me. But I suspect this is an easy problem for someone who knows the subject well.