If $(C,A)\neq (C',A')$, that is, $C\neq C'$ or $A\neq A'$, but $(T^i(C), T^{i+1}(A))=(T^i(C'), T^{i+1}(A'))$ for some $i\in \Bbb N$, that is $T^i(C)=T^i(C')$ and $T^{i+1}(A)=T^{i+1}(A')$.
Do we need $T^i(C) \to T^{i+1}(A)$ is equal to $T^i(C')\to T^{i+1}(A')$?
You may see Cohomological functors as a generalization of derived functors (Ext, Tor, Local Cohomology,...) and the Cohomology functor $H^\bullet$ which is natural - which means axiom (ii) of Cohomological Funtor. In the literature Cohomological Functors are also called $\delta$-functors.
The answer to your question is: not necessarily. The commutative diagram involving $\delta$'s comes from a commutative diagram with short exact sequences, thus in order to get $T^i(C)\rightarrow T^{i+1}(A)$ equals to $T^i(C')\rightarrow T^{i+1}(A')$ you need to have a commutative diagram like item (ii).