By the connection between Suslin trees and diamond, it seems intuitive that they are oraclecc. However, as claimed in the previous post, it is also possible to construct an oracle $\bar{M}$ for which any Suslin tree fails to be $\bar{M}$cc.

Proof Let $\bar{M}$ be an oracle. Assume we have defined a tree $T$ up to some limit level $\delta$, i.e. we’ve defined $T \upharpoonright \delta$ such that $T \upharpoonright \delta = \delta$, each node splits and all branches extend to height $\delta$.
Now we want to use the oracle to decide which (countably many) branches to extend to level $\delta$.
Let $\langle A_n : n \in \omega\rangle$ enumerate all maximal antichains in $T \upharpoonright \delta \cap M_\delta$. For each $x \in T \upharpoonright \delta$ extend $x$ to $x_n$ for $n \in \omega$ such that $x_n \geq y_n \in A_n$ and the $x_n$’s form a cofinal branch in $T \upharpoonright \delta$. Then add $x_\omega > x_n$ for all $n \in \omega$ at height $\delta$.
Continue this process for all $x \in T \upharpoonright \delta$ and then for all $\delta$ such that $T \upharpoonright \delta = \delta$. The resulting tree $T$ is Suslin by standard arguments.
Fix $\delta$ such that $T \upharpoonright \delta = \delta$ and let $A \in M_\delta$ be a maximal antichain in $T \upharpoonright \delta$. Assume that $A$ is not maximal in $T$ and so choose a $t \in T$ which witnesses this.
Now $t \geq \delta$ and so let $s \leq_T t$ be the unique node of $T$ at height $\delta$. By construction $s$ must be above some element of $A$, contradiction.
$\Box$

Proof Start with an oracle $\bar{N}$ and let $\bar{M} \triangleright \bar{N}$ (see part 1 for details) such that $T \upharpoonright (\delta + 1) \in \bar{M}$ and for each $x \in T$ at level $\delta$ let $\{y \in T : y < x\} \in M_\delta$.
Let $\delta$ be such that $T \upharpoonright \delta = \delta$ and choose $x \in T$ at level $\delta$. The set $S = \{y \in T \upharpoonright \delta : y \nleq x\}$ is dense in $T \upharpoonright \delta$ and $S \in M_\delta$. Let $A \subseteq S$ be a maximal antichain such that $A \in M_\delta$. However, $A$ is not maximal in $T$ as $x$ is incompatible with $A$.
$\Box$
It's still an open question whether all Suslin trees are oraclecc for some oracle, but the above claims are enough to show that uberoraclecc is strictly stronger than oraclecc.