(This is only a partial answer and I suspect it only confirms what you already knew but it was an interesting exercise for me and I thought I'd write it up.)
I will use the following interpretation of your question:
Let $A$ be an integral domain whose fraction field $K$ is perfect. For a matrix $M \in \mathrm{Mat}_{n \times n}(A)$, let $$ M = M_{\mathrm{ss}} + M_{\mathrm{n}} $$ be its Jordan decomposition over $K$ (so $M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(K)$). We may ask whether $M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(A)$.
Suppose $A$ is normal. Let $p_{M} \in A[t]$ be the characteristic polynomial of $M$. Let $\overline{K}$ be an algebraic closure of $K$, let $\overline{A}$ be the integral closure of $A$ in $\overline{K}$, and let $\lambda_{1},\dotsc,\lambda_{r} \in \overline{K}$ be the distinct roots of $p_{M}$. Since $\overline{A}$ is normal, we in fact have $\lambda_{i} \in \overline{A}$. If every pairwise difference $\lambda_{i} - \lambda_{j}$ (with $i \ne j$) is invertible in $\overline{A}$, then $M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(A)$.
Remark: Your condition that $p_{M}$ divides a power of a polynomial with invertible discriminant is nicer because it makes sense over any ring, but if $A$ is a normal domain, then I think it implies my condition that $\lambda_{i} - \lambda_{j}$ is invertible. It's not clear to me whether the converse is true if $A$ is a domain which is not a UFD.
Proof: Let $K'/K$ be a finite Galois extension over which $p_{M}$ splits, and let $A'$ be the integral closure of $K$ in $K'$. Since $A$ is normal, the inclusion $A \subseteq K \cap A'$ is an equality. Thus we may reduce to the case when $p_{M}$ splits into linear factors over $A$, say $p_{M}(t) = (t-\lambda_{1})^{n_{1}} \dotsb (t-\lambda_{r})^{n_{r}}$. The usual proof of Jordan-Chevalley decomposition for perfect fields (e.g. [2, Section 4.2]) says that if $p \in A[t]$ is any polynomial with $p-\lambda_{i} \in (t-\lambda_{i})^{n_{i}}$ for all $i$, then $M_{\mathrm{ss}} = p(M)$ and $M_{\mathrm{n}} = M-p(M)$. (Here often it is assumed that also $p \in (t)$, but if I understand correctly it is only used to show that Jordan decomposition is behaves well under passing to $M$-invariant subspaces, see part (c) of the Proposition in [2, Section 4.2].) Since by assumption the pairwise differences $\lambda_{i}-\lambda_{j}$ are invertible, the polynomials $(t-\lambda_{i})^{n_{i}}$ are comaximal in $A[t]$, so the Chinese remainder theorem applies and we can find such $p$.
If the pairwise differences $\lambda_{i}-\lambda_{j}$ are not all invertible, it is not necessarily true that $M_{\mathrm{ss}},M_{\mathrm{n}} \in \mathrm{Mat}_{n \times n}(A)$. Here is an example following [1]. Let $a,b \in A$ and set $$ M_{(a,b)} := \begin{bmatrix} & & a^{2}b \\ 1 & & -(a^{2}+2ab) \\ & 1 & 2a+b \end{bmatrix} $$ which is the companion matrix associated to the polynomial $(x-a)^{2}(x-b)$. Then Example 2.3, Example 4.5 of [1] says that the matrices $M_{(a,b),\mathrm{ss}},M_{(a,b),\mathrm{n}}$ have coefficients in $A$ if and only if $a-b$ is a unit of $A$.
Let's work out the example with $A = \mathbb{Z}$ and $a = 0$ and $b = 2$. Then the eigenvalues of $$ M_{(0,2)} = \begin{bmatrix} & & \\ 1 & & \\ & 1 & 2 \end{bmatrix} $$ are $0,2$, the generalized eigenspace for $\lambda = 0$ is $\{v_{1} = [0,2,-1]^{T} , v_{2} = [2,-1,0]^{T}\}$ (with $M_{(0,2)}v_{2} = v_{1}$) and the eigenspace for $\lambda = 2$ is $\{v_{3} = [0,0,1]^{T}\}$. Set $U := [v_{1},v_{2},v_{3}]$; then $$ U^{-1}M_{(0,2)}U = \begin{bmatrix} 0 & 1 & \\ & 0 & \\ & & 2 \end{bmatrix} $$ is a Jordan normal form of $M_{(0,2)}$, and $$ M_{(0,2),\mathrm{ss}} = U \begin{bmatrix} 0 & & \\ & 0 & \\ & & 2 \end{bmatrix} U^{-1} = \begin{bmatrix} & & \\ & & \\ 1/2 & 1 & 2 \end{bmatrix} \qquad M_{(0,2),\mathrm{n}} = U \begin{bmatrix} & 1 & \\ & & \\ & & \end{bmatrix} U^{-1} = \begin{bmatrix} & & \\ 1 & & \\ -1/2 & & \end{bmatrix} $$ which are not contained in $\mathrm{Mat}_{3 \times 3}(\mathbb{Z})$.
References:
[1] Passi, Roggenkamp, Soriano, "Integral Jordan decomposition of matrices", Linear Algebra and its Applications, vol. 355 (2002), 241 -- 261 (link)
[2] Humphreys, Introduction to Lie Algebras and Representation Theory, Springer Graduate Texts in Mathematics, vol. 9 (1972)
