In order to solve the counting problem, it is easiest to use the principle of inclusion exclusion. I will solve the complementary problem, of counting orderings without any adjacent same-colored plates.
As you said, the number of total orderings is $12!/2^6$. From this count, for each color, we will subtract the orderings where the plates of that color are adjacent. For example, to count the orderings where the two blue plates are together, you treat those two blue plates as a single unit, then you permute the eleven units in $11!$ ways, and finally divide by $2^5$ to account for each of the other colors having two identical plates. So, there are $11!/2^5$ ways for the blue plates to be together; we must subtract this for each color, so we subtract $6\times 11!/2^5$.
However, stacks with two same-colored adjacent plates will be doubly subtracted in the last paragraph, so we must add back in the number of such doubly counted arrangements to compensate. There are $\binom 62$ pairs of colors, and each pair has $10!/2^4$ arrangements where both colors have their plates together, so we add in $\binom 62\times 10!/2^4$. The principle of inclusion exclusion says we must continue in this fashion, next subtracting the stacks where three colors are together, then adding in the stacks where four colors are together, etc.
The final answer for the number of arrangements where no colors are together is
$$
\frac{12!}{2^6}-\binom 61\frac{11!}{2^5}
+\binom 62\frac{10!}{2^4}-\binom 63\frac{9!}{2^3}
+\binom 64\frac{8!}{2^2}-\binom 65\frac{7!}{2^1}
+\binom 66 6!\\=2631600.
$$
Therefore, the probability of no same-colored pairs is $\frac{2631600}{12!/2^6}\approx 0.3516$. In particular, about $65\%$ of the time, you should see at least one same-colored adjacent pair. This agrees with your observations.
Generalizing this logic, when you have $n$ pairs of plates stacked in a random order, the probability that there are no same-colored adjacent plates is
$$
\sum_{k=0}^n (-1)^k\binom nk \frac{(2n-k)!\cdot 2^k}{(2n)!}.
$$
As $n$ approaches infinity, this probability approaches $1/e\approx 0.367879.$
Poisson Approximation
A powerful tool for many combinatorial problems is Poisson approximation. The general idea is that when you have lots of rare events, and those events are approximately independent*, then the number of events which occur which be approximately Poisson distributed.
Specifically, suppose there $n$ events $E_1,\dots,E_n$, with probabilities given by $\mathbb P(E_i)=p_i$ for each $i$. Let $X$ be the random variable equal to number of events which occur. Let $\lambda=\mathbb E[X]=\sum_{i=1}^n p_i$, and let $Z$ be Poisson distributed with parameter $\lambda$. If Poisson approximation works, then $X$ will be approximately distributed like $Z$.
In your case, we have six events, one for each color, corresponding to that pair of plates of that color being adjacent. The probability of each event is $11/\binom{12}2=1/6$, so Poisson approximation requires $\lambda=6\times 1/6=1$. Then, we conclude as follows:
$$
\mathbb P(\text{no colors together})=P(X=0)\approx P(Z=0)=e^{-\lambda}=e^{-1}\approx 0.367879.
$$
This explains the appearance of $1/e$. The exact answer was $P(X=0)\approx 0.3516$, so Poisson approximation provided an answer with a $5\%$ relative error, which is not too bad for a back-of-the-envelope calculation. Notice how much time this saves compared to doing the full inclusion-exclusion method.
Following the same reasoning when there are $n$ pairs of plates, you get the same approximation of $1/e$ for the probability of no plates being together.
*When I say "approximately independent", I am being vague. See Two Moments Suffice by Arratia, Goldstein, and Gordon for a precise formulation of approximate independence. https://projecteuclid.org/journals/annals-of-probability/volume-17/issue-1/Two-Moments-Suffice-for-Poisson-Approximations--The-Chen-Stein/10.1214/aop/1176991491.full.
