For Q1 the problem is that one invariant of the matrix is the (isomorpism class
of the) cokernel and any $\mathbb Z[t]$-module generated by $n$ elements and
$n$-relations appears as such an invariant. There simply are too many modules
over a $2$-dimensional ring such as $\mathbb Z[t]$.
As for Q2 you cannot really hope for a classifiction even for the conjugacy classes
of matrices of finite order. In fact already for matrices of order $p^m$, $p$ prime
and $m>2$, the problem is wild which essentially means that any complete
classification is hopeless.
Finally (as far as these comments go) for the part of Q3 where the characteristic
polynomial is the product of two distinct irreducible polynomials $f$ and $g$:
If $M$ is $\mathbb Z^n$ as a module of $\mathbb Z[t]$ through the matrix, then
we have a direct sum part $M'\bigoplus M''\subseteq M$, where $M'$ is the
annihilator of $f$ and $M''$ of $g$. Hence, $M'$ and $M''$ are given by ideals
as in Latimer-McDuffee and can be considered classified. Then $M$ is given by a
submodule $\overline M$ of $M'\bigotimes\mathbb Q/\mathbb Z\bigoplus
M''\bigotimes\mathbb Q/\mathbb Z$. As $M'$ and $M''$ are the kernels of
multiplication by $f$ resp.\ $g$ we also get that $\overline
M\cap M'\bigotimes\mathbb Q/\mathbb Z=0$ and the same for $M''\bigotimes\mathbb
Q/\mathbb Z$. Hence $\overline M$ is the graph of an isomorphism between
submodules $\overline M'\subseteq M'\bigotimes\mathbb Q/\mathbb Z$ and
$\overline M''\subseteq M''\bigotimes\mathbb Q/\mathbb Z$. This means that
$\overline M'$ and $\overline M''$ are killed by both $f$ and $g$ and as they
are relatively prime, the kernel of $g$ on $M'\bigotimes\mathbb Q/\mathbb Z$ (as
well as the kernel of $f$ on $M''\bigotimes\mathbb Q/\mathbb Z$) are finite (and
usually quite small). Hence one can (in principle) determine the possible
$\overline M$ and they have to be considered modulo automorphisms of $M'$ and
$M''$ which are given by units in their endomorphism rings (which are overorders
of $\mathbb Z[t]/(f)$ resp. $\mathbb Z[t]/(g)$).
This sometimes works very well. For instance this is exactly one way of doing
the classification of matrices of order $p$ (where $f=t-1$ and
$g=t^{p-1}+\cdots+t+1$). On the other hand I am pretty sure it is as
hopeless as the general conjugacy problem for general $f$ and $g$.
There do exist pairs of finite unital rings whose additive structures
are isomorphic and whose multiplicative structures are isomorphic,
yet the rings themselves are not isomorphic.
To see this, let $\mathbb F$ be a field and let $X = \{x_1,\ldots, x_n\}$
be a set of variables. The polynomial ring $\mathbb F[X]$
is graded by degree
$$
\mathbb F[X] = H_0\oplus H_1\oplus H_2\oplus\cdots.
$$
Let $Q(x_1,\ldots,x_n)$ be a quadratic form over $\mathbb F$.
Let
$$I = \mathbb F\cdot Q(X)\oplus H_3\oplus H_4\oplus\cdots$$
be the
ideal generated by $Q(X)$ and the homogeneous components
of degree at least $3$.
Let $S_{\mathbb F,Q}$ denote the $\mathbb F$-algebra
$\mathbb F[X]/I$. It is a commutative, local ring, which encodes
properties of the quadratic form $Q$.
Two quadratic forms $Q_1$ and $Q_2$ are equivalent
if they differ by an invertible linear change of variables.
Claim.
Let $\mathbb F$ be a finite field of odd characteristic $p$.
Let $Q_1(x_1,\ldots,x_n)$ and $Q_2(x_1,\ldots,x_n)$ be
nonzero quadratic forms over $\mathbb F$.
$S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ have isomorphic
$\mathbb F$-space structures.
If $n>4$ and $Q_1$ and $Q_2$ are nondegenerate, then
$S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ have
isomorphic multiplicative monoids.
$S_{\mathbb F,Q_1}\not\cong S_{\mathbb F,Q_2}$ as $\mathbb F$-algebras,
unless $Q_1$ is equivalent to a nonzero scalar multiple of $Q_2$.
Proof. Exercise! \\
So let $\mathbb F = \mathbb F_3$ be the $3$-element field.
It is known that over a finite field of odd characteristic
the quadratic forms are classified by the dimension
and by the determinant of the form modulo squares.
The determinant of
$$
Q(x_1,\ldots,x_n)=a_1x_1^2+a_2x_2^2+\cdots+a_nx_n^2
$$
is $a_1\cdots a_n$. If
$\alpha\in \mathbb F_3^{\times}=\{\pm 1\}$,
then $\alpha\cdot Q$ has determinant
$\alpha^n a_1\cdots a_n=(\pm 1)^n a_1\cdots a_n$.
If $n$ is even, then the determinants of $Q$ and $\alpha\cdot Q$ will be equal,
so $Q$ will be equivalent
to $\alpha\cdot Q$ for every $\alpha\in \mathbb F_3^{\times}$.
This implies that, when working over $\mathbb F_3$ in an even dimension,
if $Q_1$ is not equivalent to $Q_2$, $Q_1$ will also
not be equivalent to any nonzero scalar
multiple of $Q_2$.
In particular, no scalar multiple of
$$
Q_1 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2,
$$
is equivalent to
$$
Q_2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 - x_6^2
$$
over $\mathbb F_3$. For these forms we have
that $S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ are nonisomorphic
finite unital rings with isomorphic additive and multiplicative structures.
(These rings have size $3^{27}$.)
Minor side comment 1: If you allow nonunital rings, there is a pair of nonisomorphic $8$-element rings whose additive and multiplicative structures are isomorphic.
Minor side comment 2:
The solution to the exercise above (that is, the proof of the Claim)
can be found here.
Best Answer
Your question is equivalent to whether the category $\mathcal{E}$ of pairs $(V,f)$ consisting of a finitely generated free (right) $R$-module and an endomorphism $f$ of $V$ is a Krull-Schmidt category, i.e., an additive category where every object decomposes as a direct sum of finitely many indecomposable objects and the decomposition is unique up to isomorphism and reordering. (The category $\cal E$ is also equivalent to the category of right $R[t]$-modules which are f.g. free $R$-modules, and it is rarely Krull-Schmidt, but I won't use this point of view.)
Here is one possible counterexample, phrased using the category $\mathcal{E}$ of pairs $(V,f)$ above: Take $R$ to be a Dedekind domain admitting a non-free rank-$1$ projective module $L$ such that $L\oplus L\cong R\oplus R$ (equivalently, $L$ represents an element of order $2$ in the Picard group of $R$). Let $f_L : L^2\to L^2$ be defined by $f_L(x,y)=(0,x)$, and define $f_R:R^2\to R^2$ similarly. The isomorphism $L\oplus L\cong R\oplus R$ gives rise to an isomorphism $$(L^2,f_L)\oplus (L^2,f_L) \cong (R^2,f_R)\oplus (R^2,f_R)$$ in $\cal E$. One readily checks that ${\rm End}_{\cal E}(L^2,f_L)\cong R[\epsilon|\epsilon^2=0]$ and ${\rm End}_{\cal E}(R^2,f_R)\cong R[\epsilon|\epsilon^2=0]$, so the endomorphism rings of $(L^2,f_L)$ and $(R^2,f_R)$ contain no nontrivial idempotents. This means that these objects are indecomposable in $\cal E$. On the other hand, $(L^2,f_L)\ncong (R^2,f_R)$ because $\ker f_L\cong L\ncong R\cong \ker f_R$. Consequently, the monoid $(\cal E/\cong, \oplus)$ (which is isomorphic to $(J(R)/\sim, \oplus)$ in your question) is not a free abelian monoid (because $2x=2y$ implies $x=y$ in a free abelian monoid).
One the other hand, a sufficient (but not necessary) condition for the category $\cal E$ to be Krull-Schmidt is that the endomorphism ring of every object $(V,f)$, i.e., the centralizer of $f$ in ${\rm End}_R(V)\cong {\rm M}_n(R)$, is a semiperfect ring.
For example, if $R$ is commutative and aritinian as in Benjamin Steinberg's comment, then ${\rm End}_{\cal E}(V,f)$ will be an $R$-subalgebra of ${\rm End}_R(V)$, hence artinian, and in particular semiperfect.
The semiperfectness ${\rm End}_{\cal E}(V,f)$ for all f.g. free $V$ is actually true even if $R$ is non-commutative one-sided artinian, and even if $R$ is just semiprimary. This appears implicitly in a paper of mine (page 20 & Thm. 8.3(iii) & Remark 2.9). When $R$ is commutative noetherian and local, one can use a theorem of Azumaya (Theorem 22) and a little work to show that this property is equivalent to $R$ being henselian.