Are equivalences of categories stable under filtered colimits? Let $D \colon \mathbf{J} \to \mathbf{Cat}$ be a filtered diagram of categories and functors. It has a colimit $\mathbf{C} = \mathrm{colim}\;D$. If you replace the diagram by a naturally isomorphic one $D' \colon \mathbf{J} \to \mathbf{Cat}$, then the colimit
$\mathbf{C'} = \mathrm{colim}\;D'$ is isomorphic to $\mathbf{C}$. What happens if you replace the diagram by a naturally equivalent one $D'' \colon \mathbf{J} \to \mathbf{Cat}$, is the colimit $\mathbf{C''} = \mathrm{colim}\;D''$ still equivalent to $\mathbf{C}$?
(I'm interested in $\mathbf{Cat}$, or in fact the category $\mathbf{MonCat}$ of monoidal categories, but suspect that there is a general statement in weak 2-categories. Any references are appreciated.)
 A: With regards the question about MonCat, I will assume that you are taking strong monoidal functors as morphisms.  (If you are dealing with strict morphisms things are straightforward.)
Then MonCat is an accessible category with filtered colimits and U:MonCat--->Cat preserves them.  Since, viewing it as a 2-functor, it also reflects equivalences this reduces the MonCat question to the Cat one.
More generally, if T is a finitary flexible (aka cofibrant) 2-monad on Cat, the 2-category of algebras and pseudomaps T-Alg is accessible with filtered colimits and U:T-Alg-->Cat preserves them.  This is a special case of Corollary 7.3 of my paper  https://arxiv.org/abs/2003.06375.  Flexibility or cofibrancy of the 2-monad means that it describes categorical structures with no equations between objects (like monoidal categories but not strict monoidal categories).  For strict structures like strict monoidal categories, it is not true.
As to why filtered colimits of equivalences are equivalences in Cat, the approach via combinatorial model categories described by Simon Henry is a good general approach.  An argument avoiding model categories, though similar in spirit, is to observe that, viewed as objects of the arrow category of Cat, equivalences are finitary injectives - see Diagram 2.3 of https://arxiv.org/abs/1712.02523 - and then use that finitary injectivity classes are always closed under filtered colimits, which is easy to prove (this is an easy part of the proof of Prop 4.7 of Locally presentable and accessible categories.)
P.S. Perhaps the original result of this nature is Lemma 5.4.9 of "Accessible categories .." by Makkai and Pare which shows that filtered colimits are bicolimits in Cat.  This implies the property about equivalences.
A: 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.
A: You have to take the homotopy colimit, which is the Grothendieck construction localized at a certain class of morphisms (arguably this is more computable than the 1-categorical colimit, so this is actually nice).  It agrees up to equivalence with the ordinary 1-categorical colimit when the diagram is projectively cofibrant in the folk model structure on Cat.
So the answer is negative unless you take the appropriate 2-categorical universal construction as your definition of colimit.
However, an interesting thing to note is that if you work ∞-categorically, the (∞,1)-category of 1-categories (which is a (2,1)-category) does produce the correct answer (since in that case, all colimits are ∞-colimits).  That's also why the model category picture produces the same answer.
You can also use the Bousfield-Kan formula as another presentation of the homotopy colimit, fwiw (see here).
Edit: Here's a silly counterexample:
Consider the diagram
$$\ast \leftarrow \ast \coprod \ast \to \ast$$
where $\ast$ denotes the terminal category.  If we form the 1-categorical colimit, we clearly obtain a point.
However, this diagram is naturally equivalent to
$$J\hookleftarrow \ast\coprod\ast \hookrightarrow J$$
Where $J$ denotes the free-walking isomorphism (unique contractible groupoid with two objects).  (The transformation here is the identity on the middle term and the terminal map on the outer terms.  This is strictly natural, although its quasi-inverse is only pseudonatural).
In this case, the strict colimit is $$\Sigma (\ast \coprod \ast)=\operatorname{B}\mathbb{Z},$$ which is the correct homotopy colimit.
We can also verify that we get the same answer by doing the Grothendieck construction and localizing at the Cartesian arrows.  If we call this diagram $D$, then the Grothendieck construction $$\int D$$ is the category
$$\begin{matrix}
&&\ast&&\\
&\swarrow&&\searrow&\\
\ast && && \ast\\
&\nwarrow&&\nearrow&\\
&&\ast&&
\end{matrix}$$
Localizing at the Cartesian morphisms (which are in this case all of the nonidentity morphisms), we obtain the associated totally localized groupoid, which is equivalent to $S^1=\operatorname{B}\mathbb{Z}$.
