In order not to have to worry about size issues, I'm going to answer the following question instead:
For a (small) cardinal number $\kappa$, is the category of small categories with $\kappa$-small 2-colimits 2-cocomplete?
If you take $\kappa$ to be inaccessible, then this will correspond to your question, under a particular choice of foundations. I presume moreover that you mean "2-colimits" in the weak "up-to-equivalence" sense which the nLab uses (which 2-category theorists traditionally call "bicolimits").
The fact that the 2-category Cat of small categories is 2-cocomplete, in this sense, has been well-known to category theorists for decades. It is obvious that Cat is cocomplete as a 1-category (since it is locally finitely presentable), and since it is closed symmetric (cartesian) monoidal, it follows by general enriched category theory that it is cocomplete as a category enriched over itself. In the nLab terminology, it has all strict 2-colimits. We then observe that strict pseudo 2-limits, which are 2-limits that represent cones commuting up to isomorphism but satisfy their universal property up to isomorphism (rather than equivalence), are particular strict 2-limits. Since any strict pseudo 2-limit is also a (weak) 2-limit, Cat is 2-cocomplete.
Now as Zoran pointed out in the comments, there is a 2-monad on Cat whose algebras are categories with $\kappa$-small colimits; let us call this 2-monad $T$. The strict $T$-morphisms are functors which preserve colimits on-the-nose, while the pseudo $T$-morphisms are those which are $\kappa$-cocontinuous in the usual sense (preserve colimits up to isomorphism). Therefore, the question is whether the 2-category $T$-Alg of $T$-algebras and pseudo $T$-morphisms is 2-cocomplete.
The answer is yes: it was proven by Blackwell, Kelly, and Power in the paper "Two-dimensional monad theory" that for any 2-monad with a rank (preserving $\alpha$-filtered colimits for some $\alpha$) on a strictly 2-cocomplete (strict) 2-category, the 2-category $T$-Alg is (weakly) 2-cocomplete. The 2-monad $T$ has a rank (namely, $\kappa$, more or less), so their theorem applies. I believe this all works just as well in the enriched setting.
The category $Func(C,C)$ is very rarely braided.
It's a bit like asking "when is the endomorphism algebra of a vector space commutative?"
For example, if $C=Vect\oplus Vect$, then $Func(C,C)$ is the category of Vect-valued $2\times 2$ matrices. Its multiplication does not satisfy $x\otimes y \simeq y\otimes x$, and so it's impossible to put a braiding on it.
Now you ask:
"Do braided structures occur on endofunctor categories of monoidal categories?"
Again the answer is no.
It's a bit like asking: "But what if my vector space $V$ is equipped with an algebra structure? Could then $End(V)$ be commutative?". Clearly not. The fact that $V$ is equipped with an algebra structure has no effect whatsoever on $End(V)$.
Best Answer
For filtered diagram (as asked in the question) the answer is yes. Of course this fails for general diagram as mentioned in Harry's answer.
Of course the "equivalence" has to be implemented by a pseudo-natural equivalence $f_i:F_i \to F'_i$ otherwise it is not really an equivalence of diagram.
First a very concrete proof: one constructs "by hand" a fully faithful and essentially surjective functor:
For each object $x \in \text{colim }F_i$, one chose a representaive $(i,x_0 \in F_i)$ and one define $f(x)$ to be $(i,f_i(x_0)) \in \text{colim }F'_i $.
One then define $f$ on arrows:
given $t:x \to y$ in $\text{colim }F_i$, $t$ one chose $i$ such that $t$ is defined at level $F_i$, and which is larger than the $i_x$ and $i_y$ used to define $f(x)$ and $f(y)$, ideally, we would like to define $f(t)$ as $f_i(t)$. But the source and target are not quite right $f_i(t) : f_i(x) \to f_i(y)$, while we want a map $f_{i_x}(x) \to f_{i_y}(y)$. The trick is that the pseudo-natrulity of $(f_i)$ gives us canonical isomorphism in $F'_i$, $f_{i_x}(x) \simeq f_i(x)$ here I'm using an abuse of notation, the $x$ on the left is in $F_{i_x}$ and on the right it is in $F_i$, this iso is really just the pseudo-naturality). We define $f(t)$ as the transport of $f_i(t)$ along these isomorphisms.
We then show by usual methods that this defines a functors, that it is fully faithful and essentially surjective.
The argument can be generalized to the monoidal case by a painful treatment "by hand" of the monoidal coherence.
A more general argument:
One can show very generally that in a combinatorial model category that has a set of generating cofibrations that are between $\kappa$-presentable objects the class of weak equivalences is closed under $\kappa$-filtered colimits. (see proposition 4.1 in this paper of Raptis and Rosicky)
This applies to the folk model structure on Cat is (for $\kappa=\omega$) i.e. it shows that equivalences of categories are stable under $\omega$-filtered (i.e. filtered) colimits. The same applies for the analogue of the folk model structure on monoidal categories (the one obtained for e.g. by transfer from the folk model structure).
Well, to be precise, that does not quite give us the result we want yet: it says that the colimit of a strictly natural equivalence is an equivalence, but we want it from pseudo-natrual transformation (in the monoidal case it gives us something about strictly monoidal functors).
The trick to get the result we want it to use the notion of "flexible replacement" (as for example here ) of a diagram of categories that allows to turn a pseudo-natural transformation into a span of strictly natural transformation.
In very short, given any diagram $ F:I \to $ Cat, there is a diagram $\overline{F}:I \to $ Cat, such that pseudo-natural transformation $F \to G$ are the same as strictly natural transformation $\overline{F} \to G$ (in particular one has a strictly natural equivalence $\overline{F} \to F$ and a lax transformation $F \to \overline{F}$) and every lax transformation $F \to G$ is equivalent to the span of strict transformation $\overline{F} \to F$ and $\overline{F} \to G$ where the first one is an equivalence.
A final remarks: the first argument use the axiom of choice. The second argument avoids it by only constructing a span of equivalence. But one can also write a more natural version of the first argument by only constructing an "anafunctor" in the sense of makkai, and considering all $i$ at the same time everytime.