Yes, they are equivalent, and this is why people sometimes use $|C|$ to denote $Str(C)$.
Consider the following composite $Fun(\Delta^{op},\mathrm{Grpd}) \to Fun^{cpl, Segal}(\Delta^{op}, \mathrm{Grpd}) \to Cat_\infty \to \mathrm{Grpd}$ where the first map is the left adjoint to the inclusion of complete Segal spaces, the second is the equivalence between complete Segal spaces and $\infty$-categories, and the last one is $Str$.
I claim that this map is given by geometric realization, i.e. $\mathrm{colim}_{\Delta^{op}}$. For this, because all the maps appearing are left adjoints, it suffices to show that the composite of right adjoints is equivalent to the constant functor.
But the string of right adjoints sends an $\infty$-groupoid $X$ to the $[n]\mapsto map([n],X)$, which is constant equivalent to $X$ - indeed, because $|\Delta^{op}|$ is contractible, to show that a simplicial groupoid is constant it suffices to show that it sends all maps $[0]\to [n]$ in $\Delta^{op}$ to equivalences, but $map([0],X)\to map([n],X)$ is an equivalence, as the map of $\infty$-categories $[n]\to [0]$ induces an equivalence $Str([n])\to Str([0])$.
In particular this shows that the equivalence $map([\bullet ], X)\simeq X$ is natural in $X$ as well, as it is induced by a natural transformation in $\Delta^{op}$.
It follows that if you precompose it with the inclusion $Fun^{cpl,Segal}(\Delta^{op},\mathrm{Grpd})\to Fun(\Delta^{op},\mathrm{Grpd})$, you get exactly $\mathrm{colim}_{\Delta^{op}}$ , but because the composite $Fun^{cpl,Segal}(\Delta^{op},\mathrm{Grpd})\to Fun(\Delta^{op},\mathrm{Grpd})\to Fun^{cpl, Segal}(\Delta^{op},\mathrm{Grpd})$ is equivalent to the identity, this composite is also equivalent to $Str$, which proves the claim.
No, they are not equivalent, even for $C = Sp$.
Indeed, the category of spectra with $S^1$-action is also the category of $\mathbb S[S^1]$-modules, and is compactly generated by a single object.
On the other hand, compact objects of $Fun(\mathbb Z, Sp)$ are retracts of finite colimits of representables, and any finite set $S$ of representables cannot generate the whole thing - e.g. because if $n$ is below all the elements in $S$, then $F(n) = 0$ for any $F$ generated under colimits by $S$ . So it is not compactly generated by a single object (you have to change the proof a bit, but the same holds for $Fun(\mathbb N, Sp)$)
Best Answer
Any map $f:A\to B$ fits in a cofiber sequence of arrows $(0\to A)\to (A\to A\oplus B)\to (A\to B)$
In other words, any map is a cofiber of split inclusions. But now cofibers (as any colimit, by the Bousfield-Kan formula) can be rewritten as geometric realizations of coproducts of the involved terms, and thus we can conclude (coproducts of split inclusions are clearly split inclusions).