The answer is yes or no, depending on how the quantifiers are placed.
If the question is this: Is it true that
$\forall (a_n)\ \forall (b_n)\ \exists (c_n)\ \ \big[\sum_{k=0}^n a_k=o(b_n)\iff a_n = o(c_n)\big]$?
then the answer is yes. Everywhere here, $(a_n),(b_n),(c_n)$ are sequences of positive real numbers.
Indeed, for any such $(a_n)$ and $(b_n)$, for all $n$ just let $c_n:=na_n$ if $\sum_{k=0}^n a_k=o(b_n)$ and $c_n:=a_n$ otherwise.
More interesting is this question: Is it true that
$\forall (b_n)\ \exists (c_n)\ \forall (a_n)\ \ \big[\sum_{k=0}^n a_k=o(b_n)\iff a_n=o(c_n)\big]$?
Here the answer is no. Indeed, suppose that, to the contrary, the latter highlighted statement holds. For all natural $n$ let
$$b_n:=n\ln n.$$
Then I claim that for any $(c_n)$ such that
\begin{equation*}
\forall (a_n)\ \ \big[\sum_{k=0}^n a_k=o(b_n)\iff a_n=o(c_n)\big] \tag{0}
\end{equation*}
we have
\begin{equation*}
c_n>\sqrt n
\end{equation*}
eventually, that is, for all large enough $n$.
Indeed, suppose otherwise: that for some natural $n_1<n_2<\cdots$ and all natural $j$
\begin{equation*}
c_{n_j}\le\sqrt{n_j}. \tag{1}
\end{equation*}
For all natural $k$, let now
\begin{equation*}
a_k:=\sum_{i=1}^\infty\sqrt{n_{i^2}}\,I\{k=n_{i^2}\},
\end{equation*}
where $I\{\cdot\}$ denotes the indicator. That is, $a_k=\sqrt{n_{i^2}}$ if $k=n_{i^2}$ for some natural $i$, and $a_k=0$ otherwise. Then
\begin{equation*}
\sum_{k=0}^n a_k
=\sum_{i=1}^\infty\sqrt{n_{i^2}}\,I\{n_{i^2}\le n\}
\le\sum_{i=1}^\infty\sqrt n\,I\{i^2\le n\}\le n=o(b_n).
\end{equation*}
So, by (0), $a_n=o(c_n)$, whence
\begin{equation*}
\sqrt{n_{i^2}}=a_{n_{i^2}}=o(c_{n_{i^2}})=o(\sqrt{n_{i^2}}),
\end{equation*}
by (1). This contradiction proves that indeed $c_n>\sqrt n$ eventually.
So, letting now $a_n:=\sqrt n/\ln(n+1)$, we have the condition $a_n=o(c_n)$ satisfied. However, here
\begin{equation*}
\sum_{k=0}^n a_k\sim\tfrac23\, n^{3/2}/\ln(n+1),
\end{equation*}
which is not $o(b_n)$. Thus, the second highlighted statement is false.
