Yes, this is basically correct.
A character $\chi\colon G\to \mathbb{C}^\times$ [note it should take values in the multiplicative group of non-zero complex numbers] induces an algebra homomorphism $\tilde{\chi}\colon \mathbb{C}G\to \mathbb{C}$ given by $\tilde{\chi}:=\tilde\chi(\sum \alpha_g g)=\sum \alpha_g \chi(g)$. Moreover the kernel of $\tilde{\chi}$, $$ \ker \tilde{\chi}:=\{\lambda\in \mathbb{C}G:\tilde{\chi}(\lambda)=0\}$$ is always a proper ideal of $\mathbb{C}G$, that is not equal to $\mathbb{C}G$, by the rank-nullity theorem, for example.
Your condition that $\tilde{\chi}(p_i)=0$ for all $i=1,\ldots,n$ forces $I=(p_1,\ldots,p_n)$ to be contained in the kernel of $\tilde{\chi}$ and so not equal to $\mathbb{C}G$.
In fact the converse is also true, since every maximal ideal of $\mathbb{C}G$ is the kernel of some $\tilde{\chi}$, and so every proper ideal $I$ of $\mathbb{C}G$ is contained in some $\ker \tilde{\chi}$. Moreover for this choice of $\chi$, $\tilde{\chi}(p)=0$ for all $p\in I$ and so in particular for all $p$ in a generating set for $I$.