There are two (eventually) periodic sequences associated with a fraction of positive integers $a/b$: the sequence of digits, and the sequence of remainders obtained by performing long division.
Are the minimum period lengths of these two sequences necessarily the same?
The answer is yes, however one needs to be careful with the argument. Since each digit is determined by the previous remainder, it is tempting to conclude that if the sequence of digits repeats with some period, then so does the sequence of remainders. Despite the conclusion being correct, this argument is not complete: we have failed to account for the fact that many remainders can produce the same digit. From this, one can only deduce that the period length of the digits divides the period length of the remainders.
Let us assume without loss of generality that $0\leq a< b$.
Let $d_1, d_2, d_3 \dots$ be the sequence of digits, i.e., $a/b = 0.d_1d_2d_3\dots\, $.
And let $r_0, r_1, r_2, \dots$ be the sequence of remainders, where we have taken $r_0 = a$.
Let $P_d$ and $P_r$ be the minimal period lengths of the digits and the remainders respectively.
It follows from the long division algorithm that these two sequences are related by the following expression $10\cdot r_{k-1} = d_k\cdot b + r_k$.
Therefore, by induction since $r_0 = a$,
\[
\frac{a}{b} = \frac{d_1}{10}+\frac{d_2}{10^2}+\dots+\frac{d_k}{10^k} + \frac{r_k}{10^k\cdot b},
\]
for all $k\geq 0$. Let $T_k := r_k/b$. It follows from the above that $T_k = 0.d_{k+1}d_{k+2}d_{k+3}\dots\,$. That is, the decimal expansion of $T_k$ is precisely the tail of the decimal expansion of $a/b$, starting from position $k+1$. Hence, the sequence $T_0, T_1, T_2, \dots$ must eventually become periodic with minimal period $P_d$.
On the other hand, it is immediate from the definition that the sequence $T_0, T_1, T_2, \dots$ must also become periodic with minimal period $P_r$.
Therefore, we conclude that $P_r = P_d$, as we wanted.