$\newcommand{\al}{\alpha}
\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\varepsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\om}{\omega}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\F}{\mathcal{F}}
\newcommand{\E}{\operatorname{\mathsf E}}
\renewcommand{\P}{\operatorname{\mathsf P}}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}
\newcommand{\tf}{\widetilde{f}}$

I am interpreting the problem as follows: Assume that the $\ep_1,\dots,\ep_n,r_1,\dots,r_n$ are independent, and we need an upper bound on
\begin{equation}
E_K:=\E\Big(\Big|\sum_{i=1}^na_i\ep_i r_i\Big|^p\Big|\sum_{i=1}^nr_i=K\Big)
\end{equation}
for all values of $K$ for which this conditional expectation is defined. However (because of the symmetry), given any values of the $n$-tuple $(r_1,\dots,r_n)$, the conditional distribution of $(\ep_1r_1,\dots,\ep_n r_n)$ is the same as the unconditional distribution of $\al(\ep_1,\dots,\ep_n)$. Therefore, $E_K$ does not depend on $K$ (as long as $E_K$ is defined), and so, we have the best possible upper bound on $E_K$:
\begin{equation}
E_K=E:=\E\Big|\sum_{i=1}^n a_i\al\ep_i\Big|^p\le \E|Z|^p|\al|^p\Big(\sum_{i=1}^n a_i^2\Big)^{p/2},
\end{equation}
by Haagerup's inequality, where $Z\sim N(0,1)$.