Yes, and a more general statement is true even over general fields of characteristic $0$, according to Bourbaki's take on Cartan subalgebras (in book VII, §2 of the volume on Lie Groups and Lie Algebras). Namely, proposition 10 says that for an abelian subalgebra $\mathfrak{a} \subseteq \mathfrak{g}$ consisting of semisimple elements,
$$\lbrace \text{Cartan subalgebras of } \mathfrak{g} \text{ containing } \mathfrak{a} \rbrace = \lbrace \text{Cartan subalgebras of } \mathfrak{z}_{\mathfrak{g}}(\mathfrak{a}) \rbrace $$
($\mathfrak{z}_{\mathfrak g} =$ centraliser). But every Lie algebra has Cartan subalgebras (see e.g. corollary 1 to theorem 1 loc. cit.), in particular so does $\mathfrak{z}_{\mathfrak{g}}(\mathbb{C} J)$ in your question.
Took some work and some scouring through the literature, but we got there in the end. Both questions are answered positively — though my brain still needs some time to digest the answer for question one, and make sure that it's really really true, so, eh, approach with care.
First question:
Whenever I say parabolic here, I mean not the Knapp definition, but the one in terms of complexification, see the OP.
From Lemma in Section 3.2 from Wolf, Koranyi, we can extract the following (heavily paraphrased, but hopefully equivalent):
Let $\mathfrak{g}$ be a real semisimple Lie algebra and $\mathfrak{q}$ a parabolic subalgebra. Then there is some Cartan decomposition
$$\mathfrak{g} = \mathfrak{t} \oplus \mathfrak{p} $$
some maximally noncompact ("maximally split") Cartan subalgebra $\mathfrak{h} = \mathfrak{t} \oplus \mathfrak{a}$ of $\mathfrak{g}$, where
$$\mathfrak{t} \subset \mathfrak{k} \text{ (the "totally nonsplit" part)}, \quad \mathfrak{a} \subset \mathfrak{p} \text{ (the "totally split" part) },$$
a subspace $\mathfrak{a}' \subset \mathfrak{a}$ and a choice of positive roots $P$ in the restricted root space decomposition of $(\mathfrak{g}, \mathfrak{a}')$ so that
$$\mathfrak{q} = \mathfrak{g}_0 \oplus \bigoplus_{\alpha \in P} \mathfrak{g}_\alpha, \quad
\mathfrak{g}_\alpha = \{x \in \mathfrak{g} : [a,x] = \alpha(a) \cdot x \quad \forall a \in \mathfrak{a}'\}.
$$
(Careful: In the source, the root space decomposition is carried out in the complexification $\mathfrak{g}_\mathbb{C}$, but we can also carry it out in the real setting, since the ad-action of elements in $\mathfrak{a}' \subset \mathfrak{a}$ is real diagonalizable. This is always necessary for the restricted root space decomposition.)
Very verbose, but in the end, in the above notation, every parabolic subalgebra contains $Z_\mathfrak{k}(\mathfrak{a}) \oplus \mathfrak{a}$ in the $\mathfrak{g}_0$-component and some choice of positive restricted roots of $\mathfrak{a}$ in the $\bigoplus_{\alpha \in P} \mathfrak{g}_\alpha$-component. Hence every parabolic subalgebra contains some subalgebra of the form $\mathfrak{m} \oplus \mathfrak{a} \oplus \mathfrak{n}$. And indeed, the subalgebras $\mathfrak{m} \oplus \mathfrak{a} \oplus \mathfrak{n}$ are parabolic, since their complexification contains a Borel algebra associated to the complexification of the Cartan subalgebra $\mathfrak{a} \oplus \mathfrak{t}$.
Hence, Knapp's minimal parabolic subalgebras are exactly the minimal parabolic subalgebras in the usual sense.
Second question:
In Bourbaki, Chapter VIII, Exercise 3a for §5, we learn:
If $\mathfrak{q}, \mathfrak{p}$ are parabolic subalgebras of a semisimple (real or complex) Lie algebra $\mathfrak{g}$, and $\mathfrak{q} \subset \mathfrak{p}$, then the radical of $\mathfrak{p}$ is contained in the radical of $\mathfrak{q}$.
And in Bourbaki, Chapter VIII, §10, Corollary 2, we learn:
Every subalgebra $\mathfrak{n}$ of a (real or complex) Lie algebra $\mathfrak{g}$, consisting only of nilpotent elements of $\mathfrak{g}$, is contained in the nilradical of a parabolic subalgebra $\mathfrak{q}$.
As a corollary of the two: Given a subalgebra $\mathfrak{n} \subset \mathfrak{g}$, contained in the radical of some parabolic subalgebra $\mathfrak{q}$. But then there is some minimal parabolic $\mathfrak{q}_0 \subset \mathfrak{q}$ with $\mathfrak{n} \subset \text{rad}(\mathfrak{q}) \subset \text{rad}(\mathfrak{q}_0)$.
Wolf, J. A.; Koranyi, A., Generalized Cayley transformation of bounded symmetric domains, Am. J. Math. 87, 899-939 (1965). ZBL0137.27403.
Bourbaki, Nicolas, Elements of mathematics. Lie groups and Lie algebras. Chapters 7 and 8, Berlin: Springer (ISBN 3-540-33939-6). 271 p. (2006). ZBL1181.17001.
Best Answer
The first assertion is true and actually the only possibility (up to direct product of both $\mathfrak{g}$ and $\mathfrak{h}$ by some other semisimple algebra) is when $\mathfrak{g}$ is $\mathfrak{sl}_2$.
(Edit: switched to an algebraic proof)
In $\mathfrak{sl}_2$ it is easy to check that all codimension 1 subalgebras are conjugate to the Borel (parabolic) one.
It is enough to prove that if $\mathbf{g}$ is simple of rank $\ge 2$, then it has no codimension 1 subalgebra $\mathfrak{h}$.
Choose a Cartan subalgebra $\mathfrak{h}_0$ of $\mathfrak{h}$. It induces a grading $(\mathfrak{g}_\alpha)$ of $\mathfrak{g}$, which has to be a quotient of its own Cartan grading.
If $\mathfrak{h}_0=\mathfrak{g}_0$, then $\mathfrak{h}_0$ is a Cartan subalgebra of $\mathfrak{g}$, so $(\mathfrak{g}_\alpha)$ is the Cartan grading of $\mathfrak{g}$. In this case, it follows that $\mathfrak{h}$ is a graded subalgebra containing $\mathfrak{g}_0$, so there exists a nonzero root $\alpha$ such that $\mathfrak{h}=\bigoplus_{\beta\neq\alpha}\mathfrak{g}_\beta$. Using that $\mathfrak{g}$ has rank $\ge 2$ and is simple, there exist two nonzero roots summing to $\alpha$, and this implies that such $\mathfrak{h}$ is not a subalgebra, contradiction.
Next, if $\mathfrak{h}_0\neq\mathfrak{g}_0$, then being a quotient of the Cartan grading $(\mathfrak{g}_{(\gamma)})$ of $\mathfrak{g}$, we have $\mathfrak{g}_0$ reductive and containing a Cartan subalgebra of $\mathfrak{g}$. If $\mathfrak{g}\neq\mathfrak{g}_0$, we can argue as follows: $\mathfrak{g}_0$ is the sum of $\mathfrak{g}_{(\gamma)}$ where $\gamma$ ranges over some proper subspace $M$ of the space of roots. Since $\mathfrak{g}$ is simple, the set of roots $\gamma$ not in $M$ generates the space of roots (the set of roots is not contained in the union of two proper subspaces), and for each such $\gamma$, we have $\mathfrak{g}_{(\pm\gamma)}\in\mathfrak{h}$ and hence $h_\gamma\in\mathfrak{h}$. Hence $\mathfrak{h}$ contains a Cartan subalgebra of $\mathfrak{g}$, and this implies $\mathfrak{h}_0=\mathfrak{g}_0$ contradiction.
So we have $\mathfrak{g}=\mathfrak{g}_0$: in particular $\mathfrak{h}=\mathfrak{h}_0$. Hence $\mathrm{ad}(h)$ is nilpotent for every $h\in\mathfrak{h}=\mathfrak{h}_0$. Letting $\mathfrak{c}$ be a Cartan subalgebra of $\mathfrak{g}$, this implies that every element of $\mathfrak{h}\cap\mathfrak{c}$ is in the kernel of every root. Since the intersection of kernels of roots is zero in a semisimple Lie algebra, this forces $\mathfrak{c}$ to have dimension $\le 1$. Hence $\mathfrak{g}$ has rank $\le 1$, a contradiction.
Edit: I was a bit frustrated to make such a proof for such a weak result, but indeed it adapts to the following more natural and stronger (and classical) statement:
Lemma: let $\Phi$ be an irreducible root system in dimension $r\ge 1$. Then $\Phi$ is not contained in the union of two proper subspaces.
This follows from:
Sublemma: let $\Phi$ be a root system in dimension $r$ (not necessarily generating). Suppose that $\Phi\subset V_1\cup V_2$ where $V_i$ are subspaces. Then there exist subsets $\Phi_1,\Phi_2$ such that $\Phi_i\subset V_i$, $\Phi_1\cup\Phi_2=\Phi$, and $\langle\Phi_1,\Phi_2\rangle=0$.
Proof of sublemma. This is vacuously true in dimension $0$, and more generally if $V_2=V$. In dimension $r\ge 1$, write $\Psi_1=\Phi\smallsetminus V_2$, $\Psi_2=\Phi\smallsetminus V_1$, and $\Psi_{12}=\Psi\cap V_1\cap V_2$. Clearly $\Psi$ is the disjoint union $\Psi_1\sqcup\Psi_2\sqcup\Psi_{12}$. Also, $\Psi_1,\Psi_2$ are orthogonal: indeed otherwise, we can find $\alpha\in\Psi_1$, $\beta\in\Psi_2$ with $\langle\alpha,\beta\rangle<0$, so $\alpha+\beta\in\Phi$ and this is a contradiction because $\alpha+\beta$ belongs to neither $V_1$ nor $V_2$.
Next, we consider the subspace $V_2$, and its two subspaces $W_1=V_1\cap V_2$, and $W_2$ the orthogonal of $V_1$ in $V_2$, and $\Phi'=\Phi\cap V_2=\Psi_2\sqcup \Psi_{12}$, with $\Psi_2\subset W_2$ and $\Psi_{12}\subset W_1$. We argue by induction inside $V_2$ (the trivial case $V_2=V$ being excluded), to infer that we can write $\Phi\cap V_2=\Phi'_1\cup\Phi'_2$ with $\langle\Phi'_1,\Phi'_2\rangle =0$ and $\Phi'_i\subset W_i$. Then $\Phi=\Psi_1\sqcup\Phi'_1\sqcup\Phi'_2$, with $\Phi'_2\subset V_2$ orthogonal to $\Psi_1\sqcup\Phi'_1\subset V_1$. This finishes the induction.$\Box$
Now let us proceed to the proof of the result. It's an adaptation of the previous proof. Only the first case requires a modification, which is the reason for the above lemma. Namely, let $\mathfrak{h}$ have codimension $<r$ and suppose, with the previous notation that $\mathfrak{h}_0=\mathfrak{g}_0$. In this case the grading is the Cartan grading of $\mathfrak{g}$, so $\mathfrak{h}=\mathfrak{g}_0\oplus\bigoplus_{\alpha\in\Phi\smallsetminus F}\mathfrak{g}_\alpha$, where $F$ is a subset of the root system $\Phi$ of $\mathfrak{g}$, of cardinal $<r$.
Let $V_1$ be the subspace spanned by $F$ (a proper subspace of $\mathfrak{g}_0^*$). Fix a root $\alpha\in F$, and $V_2$ its orthogonal. Then by the lemma, there exists a root $\beta\notin V_1\cup V_2$. Let $P$ be the plane generated by $\alpha$ and $\beta$. So $\Phi\cap P$ is an irreducible root system in $P$, and we can find in $P$ two roots, not collinear to $\alpha$, and avoiding the $V_2\cap P$ (which has dimension $\le 1$), with negative scalar product and summing to $\alpha$. This shows that $\mathfrak{g}_\alpha\subset\mathfrak{h}$, a contradiction.
In the other case $\mathfrak{g}_0\neq\mathfrak{h}_0$, we almost only need to copy the previous proof.