For any three integers $m,n,b$ such that
$$m \ge 1,\quad n, b > 1\quad\text{ and }\quad\gcd(m,n) = \gcd(m,b) = \gcd(n,b) = 1$$
Let $\displaystyle\;\tau\left(\frac{m}{n}, b\right)\;$ be the "period" of the expansion of $\displaystyle\;\frac{m}{n}\;$ in base $b$.
Let $\lambda = \lambda(n)$ be the value of Carmichael function for $n$, we have
$$\gcd(b,n) = 1 \quad\implies\quad b^\lambda \equiv 1 \pmod n \quad\iff\quad n\;|\;b^\lambda - 1$$
Let $\displaystyle\;k = \frac{b^\lambda - 1}{n} \in \mathbb{Z}_{+}\;$, we have
$$\frac{m}{n} = \frac{km}{b^\lambda - 1} = km\sum_{\ell=1}^\infty b^{-\ell\lambda}
\quad\implies\quad\left.\tau\left( \frac{m}{n}, b \right) \right| \lambda.$$
This means $\lambda(n)$ is an upper bound for
$\displaystyle\;\tau\left(\frac{m}{n}, b\right)\;$
for any choice of $m$ and $b$.
Update
In general, let $\mathbb{Z}_n^\times$ be the multiplicative subgroup of $\mathbb{Z}/\mathbb{nZ}$ consists of ( equivalent classes of ) integers relative prime to $n$. Let $\mu = \text{ord}_n(b)$ be the order of $b$ in $\mathbb{Z}_n^\times$
(also known as multiplicative order or haupt-exponent of $b \pmod n$).
Using same argument as above, we get
$$n | b^\mu - 1 \quad\implies\quad
\left.\tau\left( \frac{m}{n}, b \right) \right| \text{ord}_n(b)\tag{*1a}$$
On the other direction, let $\displaystyle\;\nu = \tau\left( \frac{1}{n}, b \right)\;$. Being a period for the expansion of $\displaystyle\;\frac{1}{n}\;$ in base $b$, we can
find integers $\alpha, A, B$ such that
$$\frac{1}{n} = \frac{1}{b^\alpha}\left[ A + \frac{B}{b^\nu - 1}\right], \quad
\alpha \ge 0, A \ge 0, b^\nu - 1 > B > 0$$
It is clear $\displaystyle\;A = \left\lfloor\frac{b^\alpha}{n}\right\rfloor$. Introduce $C$ such that $b^\alpha = A n + C$, we can rewrite above condition as
$$\frac{C}{n} = \frac{B}{b^\nu - 1}\quad\iff\quad Bn = C(b^\nu - 1)
\quad\implies\quad n | C(b^\nu - 1)
$$
Since $\gcd(C,n) = \gcd(b^\alpha - An, n) = \gcd(b^\alpha,n) = 1$, this leads to
$$n | b^\nu - 1
\quad\implies\quad
\left.\tau\left(\frac{1}{n},b\right)\right| \text{ord}_n(b)\tag{*1b}$$
Combine $(*1a)$ and $(*1b)$, we get
$$\tau\left(\frac{1}{n},b\right) = \text{ord}_n(b)
\quad\implies\quad
\max\left\{\; \tau\left( \frac{m}{n}, b \right) : m \ge 1 \;\right\} = \text{ord}_n(b)
$$
i.e. $\text{ord}_n(b)$ is the least upper bound of $\displaystyle\;\tau\left(\frac{m}{n},b\right)\;$ when we fix $b$ and allow $m$ to vary.
Let $n = p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r}$ be the factorization of $n$ into its prime factors. We know as a group, $\mathbb{Z}_n^\times$ is isomorphic to the direct product of
its prime components.
$$\mathbb{Z}_n^\times =
\mathbb{Z}_{p_1^{e_1}}^\times \times
\mathbb{Z}_{p_2^{e_2}}^\times \times
\cdots
\mathbb{Z}_{p_r^{e_r}}^\times$$
The carmichael function of $n$ can be expressed in terms of its prime components
$$\lambda(n) = \text{lcm}( \lambda( p_1^{e_1} ), \lambda( p_2^{e_2} ), \cdots, \lambda( p_r^{e_r} ) )$$
We also know inside each $\mathbb{Z}_{p_i^{e_i}}^\times$, there is a element whose order equal to $\lambda( p_i^{e_i} )$. We can use
Chinese remainder theorem to assemble these elements into an element $a \in \mathbb{Z}_{n}^\times$ such that
$$ \text{ord}_n(a) = \lambda(n)
\quad\implies\quad\tau\left(\frac{1}{n}, a \right) = \lambda(n)
$$
What this means is $\lambda(n)$ is not just a upper bound for $\displaystyle\;\tau\left(\frac{m}{n},b\right)\;$, it is the least upper bound when we allow both
$m$ and $b$ to vary.
Conclusion -
If we don't impose more constraint on $b$, $\lambda(n)$ is the best bound we can have.
Update 2
About the question what sort of bound we will have when $b = 3$,
I have performed some numerical experiment. The result
indicate the period
$$\tau_n \stackrel{def}{=} \tau(\frac{1}{n},3) = \text{ord}_n(3)
=\text{lcm}\left(
\text{ord}_{p_1^{e_1}}(3), \text{ord}_{p_2^{e_2}}(3),\ldots,
\text{ord}_{p_r^{e_r}}(3)
\right)
$$
won't be anything simple that depends only on magnitude of $n$.
Instead, it depends strongly on how $n$ is factorized into its prime factors (as above formula suggests).
If we ignore the prime factorization of $n$, there aren't other pattern I can detect.
The only observation I have is:
It seems more than half of the time, $\tau_n$ is actually equal to $\lambda(n)$.
For $2 \le n \le 10^4$, there are $6666$ numbers relative prime to $3$.
Among these numbers, about $58.2\%$ of them have $\tau_n = \lambda(n)$
and close to $89.7\%$ of them $\tau_n \ge \frac{\lambda(n)}{5}$.
If we increase the range, say to $2 \le n \le 10^5$, the statistics is roughly
the same, $55.7\%$ for $\tau_n = \lambda(n)$ and $87.7\%$ for $\tau_n \ge \frac{\lambda(n)}{5}$.
Following is a table summarize this statistics for $n$ up to $10^4$.
$$\begin{array}{c:c:c}
\hline
s & \# \{ \lambda(n)/\tau_n = s\} & \# \{ n : \lambda/\tau_n \le s\}\\
\hline
1 & 3880 (58.21\%) & 3880 (58.21\%)\\
2 & 1285 (19.28\%) & 5165 (77.48\%)\\
3 & 369 ( 5.54\%) & 5534 (83.02\%)\\
4 & 256 ( 3.84\%) & 5790 (86.86\%)\\
5 & 190 ( 2.85\%) & 5980 (89.71\%)\\
6-10 & 353 ( 5.30\%) & 6333 (95.00\%)\\
11-15 & 165 ( 2.48\%) & 6498 (97.48\%)\\
16-20 & 37 ( 0.56\%) & 6535 (98.03\%)\\
21-25 & 27 ( 0.41\%) & 6562 (98.44\%)\\
26-50 & 57 ( 0.86\%) & 6619 (99.29\%)\\
51-75 & 19 ( 0.29\%) & 6638 (99.58\%)\\
76-100 & 10 ( 0.15\%) & 6648 (99.73\%)\\
\hline
\end{array}$$
Update 3
If you want to perform your own investigation, data of $\text{ord}_n(b)$ is available on OEIS:
In Mathematica, you can also compute $\lambda(n)$ and $\text{ord}_n(b)$ yourself
using the commands Carmichael[n] and MultiplicativeOrder[b,n].
