Yes, you are right that $L$ is not context-free.

We can use [the standard pumping lemma for context-free language](https://en.wikipedia.org/wiki/Pumping_lemma_for_context-free_languages).

For the sake of contradiction, let $p>0$ be a pumping length for $L$. Consider word $t=a^pc^pb^{p+1}b^{p+1}c^pa^p\in L$. Let $t=uvwxy$, where $|vx|\geq 1$, $|vwx|\leq p$, and $uv^nwx^ny\in L$ for all $n\ge0$.

There are two cases.

- $vx$ contain at least one $b$. Then $vx$ does not contain $a$. 
Let $s=uwy=uv^0wx^0y\in L$. However, it contain less $b$s than $t$ but no less $a$s than $t$. Note that the number of $b$s in all words in $L$ is even. That means $|s|_b\le|t|_b-2=|t|_a=|s|_a$. 

- $vx$ does not contain $b$. WLOG, suppose $vx$ contains one $a$. Let $s=uv^3wx^3y\in L$. However, $|s|_a|\ge|t|_a+2\gt|t|_b=|s|_b$.

In all cases, we can pump $t$ to $s\not\in L$, which contradicts that $p$ is a pumping length of $L$. This contradiction shows $L$ is not context-free.