No such ideal exists.
A simple Lie algebra $\mathfrak{g}$ over a field $k$ is called absolutely simple if for every algebraic extension $K\vert k$ (or equivalently: for an algebraic closure $K\vert k$) , the scalar extension $K\otimes \mathfrak{g}$ is also simple (note that it necessarily is semisimple; for an example where it is not simple, see below).
One can show that if a Lie algebra over a field $k$ is simple but not absolutely simple, it is the scalar restriction of an absolutely simple Lie algebra over some algebraic extension $K \vert k$. Actually, one can compute $K$ as the (associative) subalgebra of $End_k(\mathfrak{g})$ consisting of those elements that commute with all $ad_\mathfrak{g}(x), x \in \mathfrak{g}$. As far as I know, this was first shown by Jacobson in Duke Math. J., Volume 3, Number 3 (1937), 544-548, doi:10.1215/S0012-7094-37-00343-0, and holds for more general kinds of algebras. I wrote a little overview of that in section 4.1 of my thesis, a lot of it is now reproduced in this answer of mine.
Now to your question: Since scalar extension commutes with direct sums, and ideals of semisimple Lie algebras are direct summands, your question is equivalent to asking whether there exists a simple compact real Lie algebra $\mathfrak{g}$ which is not absolutely simple. But by the above theory, and the fact that the only proper algebraic extension of $\Bbb R$ is $\Bbb C$, the only simple but not absolutely simple real Lie algebras are: the simple complex Lie algebras considered as $\Bbb R$-algebras. The first example maybe being $\mathfrak{sl}_2(\Bbb C)$ viewed as a Lie algebra over $\Bbb R$ (six-dimensional); it is simple, but not absolutely simple, as its scalar extension $\Bbb C \otimes_{\Bbb R} \mathfrak{sl}_2(\Bbb C)$ actually is isomorphic to the sum of two copies of $\mathfrak{sl}_2(\Bbb C)$.
However, none of these scalar restrictions of simple complex Lie algebras correspond to compact Lie groups, e.g. because they obviously contain nilpotent elements.
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
For split real semisimple Lie algebras, yes. Otherwise, not necessarily.
Namely, if I'm not mistaken, in the notation of Bourbaki's Groupes et algèbres de Lie, your first ("algebraic") definition is denoted as $Aut_{e}(\mathfrak{g})$ (cf. ch. VII §3 no. 1 definition 1 and ch. VIII §5 no. 2). The second ("analytic") definition is denoted as $Int(\mathfrak{g})$ (cf. ch. III §6 no. 2 definition 2). And they prove that for a semisimple split Lie algebra over $\Bbb R$ or $\Bbb C$, we have $Aut_e(\mathfrak{g}) = Int(\mathfrak{g})$, in ch. VIII, §5 no. 5 proposition 11(iv).
$\mathfrak{g}$ being split here means that there exists a splitting Cartan subalgebra (i.e. one that consists of ad-diagonalisable instead of just ad-semisimple elements). Over $\Bbb C$, that is of course an empty condition, but over $\Bbb R$, it severely restricts the scope of the theorem, as it excludes the plethora of non-split real semisimple Lie algebras which exist.
For those, the theorem is certainly not true in general. As an extreme case, take any compact real Lie algebra, e.g. $\mathfrak{su}_n$. Those contain no nonzero nilpotent elements at all, so that $Aut_e(\mathfrak{su}_n)$ is trivial. However, if I understand this MO post correctly, we have e.g. $Aut(\mathfrak{su}_2) = Int(\mathfrak{su}_2) \simeq SU(2)/\pm I \simeq SO(3)$.
Finally, let me note that I find ch. VIII §5 of that Bourbaki volume one of the best treatises on automorphisms of simple Lie algebras, and to add to the existing ambiguity, it's actually interesting to look at a third possible group they call $Aut_0(\mathfrak g)$, which are the automorphisms which become elementary after scalar extension to an algebraic closure. This is related to the real Lie algebra $\mathfrak{sl}_2(\Bbb R)$ having a "non-inner" automorphism called $\varphi$ in this answer, even though one usually thinks of $\mathfrak{sl}_2$ as not having outer automorphisms. (This one, in this case, is in $Aut(\mathfrak g)= Aut_0(\mathfrak g)$ but not in $Aut_e(\mathfrak g)$.)