The idea you are flirting with is the ``Ratliff-Rush operation''. Given a regular ideal $J$ (which is a special case of an ideal with positive height), Ratliff and Rush define $\tilde{J} := \bigcup_{n=0}^\infty (J^{n+1} : J^n)$.
Then they show that $\tilde{J}$ is the largest ideal among ideals $I$ such that $I^n = J^n$ for $n \gg 0$.
Later authors have called $\tilde{J}$ the Ratliff-Rush ideal associated with $J$ (or sometimes [misleadingly] the Ratliff-Rush closure of $J$ -- misleading because the operation does not preserve containments of ideals, whereas a true closure operation should). In general, a Ratliff-Rush ideal is an ideal $K$ such that $K = \tilde{K}$.
Your condition can be restated as: $J \subseteq I \subseteq \tilde{J}$. You are right that the condition is stronger than $J$ being a reduction of $I$. For instance, if $R=k[x,y]$ and $J = (x^2, y^2)$, then $J = \tilde{J}$, even though $J$ is a reduction of $(x,y)^2$.
Question 2 (as well as question 1) has a negative answer whenever $R$ is a Dedekind domain, since every ideal there is integrally closed, hence a Ratliff-Rush ideal. I expect that both questions should have a positive answer in most cases, however.
A paper of Heinzer, Lantz and Shah proves the following: Let $R$ be a one-dimensional local domain. Then every ideal of $R$ is Ratliff-Rush if and only if every ideal either has principal reduction or reduction number $1$. This theorem should have bearing on your question as well. (In this situation, of course, the notions of parameter ideal and principal ideal coincide.)
For what I have said here along with other stuff you will likely find interesting, see the survey article http://www.math.purdue.edu/~heinzer/preprints/colconf10.pdf.
So, not necessarily a complete answer, but this should be enough information to lead you to as complete an answer as is available.
In addition to two good answers, maybe one case the question has a positive answer is when $(R,m)$ is a regular local ring and $f$ is a nonzero element. For a Noetherian local ring $A$ and ideal $I$ which is primary to the maximal ideal, let $e(I, A)$ denote the Hilbert-Samuel multiplicity of $A$ with respect $I$. Then $e(m/(f),R/(f)) = \hbox{ord} (f) e(m, R) = \hbox{ord} (f)$ where $\hbox{ord} (f) = \sup \{ i \mid f \in m^i \}$. Here $e(m, R) = 1$ since $R$ is a regular local ring.
The special case is when the dimension of a ring (not necessarily regular) $R$ is $1$ and $f$ is a non zero-divisor (but not a unit) of $R$. Assume that $R$ is Cohen-Macaulay (domain or reduced implies Cohen-Macaulyness in dim $1$). Then we have that length $(R/(f)) = e((f), R) \ge \hbox{ord} (f) e(m,R)$ where the last inequality follows from Theorem 14.10 in Matsumura's commutative ring theory. A positive answer to your question implies an equality in the formula and $e(m,R) = 1$. The latter condition is equivalent to $R$ being regular if $R$ is unmixed. Note that the associated graded ring is a polynomial ring over the residue field if a ring is regular. In particular, it is a domain. I believe that once $R$ is a regular local ring, then the equality follows.
I believe in Hartshorne's algebraic geometry book, a surface is a nonsingular (locally regular) projective surface over an algebraically closed field. This probably is a reason why it works well. I hope someone can add some geometric point of view.
Best Answer
I follow user @Z. M's comment.
If $M = \cup_{i=0}^\infty F_i M$ is a filtered module and $$0 \to M' \to M \to M'' \to 0$$ is a short exact sequence such that $M', M''$ have the induced filtrations, then we get a short exact sequence of Rees modules $$0 \to R_hM' \to R_hM \to R_hM'' \to 0,$$ which yields the short exact sequence $0 \to \mathrm{gr}(M') \to \mathrm{gr}(M) \to \mathrm{gr}(M'') \to 0$ after taking the Snake Lemma (since our filtration is by submodules).
Now in our situation, $A/I$ and $I$ have the induced filtrations, so we get a short exact sequence $ 0 \to \mathrm{gr}(I) \to \mathrm{gr}(A) \to \mathrm{gr}(A/I) \to 0$.