Let $\mathfrak{g}$ be a Lie algebra with $\mathfrak{a}$ a maximal commutative subalgebra. We then have the root decomposition $$\mathfrak{g}=\mathfrak{a} \oplus \bigoplus _{\alpha \in \Psi} \mathfrak{g}_\alpha$$ for a set of simple roots $\Psi \subset \mathfrak{a}^*$. A maximal solvable subalgebra $\mathfrak{b}$ of $\mathfrak{g}$ is called Borel subalgebra. (n.b. Borel algebras are always conjugated to algebras of the form $\mathfrak{a} \oplus \bigoplus _{\alpha \in \Psi^+} \mathfrak{g}_\alpha$ for a set of positive roots $\Psi^+ \subset \Psi$)
A subalgebra $\mathfrak{p} \subset \mathfrak{g}$ is called parabolic subalgebra if it contains a Borel subalgebra. I have found two different definitions for minimal parabolic subalgebra.
Onishchik, Vinberg: Lie Groups and Lie Algebras III, p. 191: $\mathfrak{q}(\emptyset)=\mathfrak{g}_0 \oplus \bigoplus _{\alpha \in \Psi^+} \mathfrak{g}_\alpha$.
Knapp: Lie Groups Beyond an Introduction, p. 270: $\mathfrak{q}=\mathfrak{b}$.
Clearly, $\mathfrak{q}=\mathfrak{b}$ is the smallest subalgebra containing $\mathfrak{b}$. Why would one want to call the bigger algebra $\mathfrak{q}(\emptyset)$ a minimal parabolic Lie algebra?
The notion of "minimal parabolic" applies, meaningfully, more generally than does the notion of "Borel subgroup/subalgebra".
It depends whether you're working over $\mathbb R$ or over $\mathbb C$, since over $\mathbb R$ a Levi-Malcev component of a minimal parabolic may be quite large. For example, in (the Lie algebra of) $O(n,1)$, there is a unique conjugacy class of parabolic subgroup (apart from the group itself), and they all have Levi-Malcev component isomorphic to $O(n)\times O(1)$. In this case, one might hesitate to call this minimal parabolic a Borel subgroup/subalgebra, because of the discrepancy between real and complex root spaces.
But for "split" or "quasi-split" groups/algebras, the minimal parabolics are indeed Borel.
Over $\mathbb C$, they are always the same.