Surely it's easier to check whether $H'$ is simply-connected by inspecting the co-root lattice...? For the example of $G_2$ containing $SO(4)$ that Allen mentions, we have a pseudo-Levi subalgebra of type $A_1\times \tilde{A}_1$ where $\{(3\alpha+2\beta),\alpha\}$ is a basis of simple roots. Now the cocharacter $(3\alpha+2\beta)^\vee=\alpha^\vee+2\beta^\vee$, so the lattice:
${\mathbb Z}(3\alpha+2\beta)^\vee+{\mathbb Z}\alpha^\vee = {\mathbb Z}\alpha^\vee+{\mathbb Z}(2\beta^\vee)$
is of index two in the cocharacter lattice ${\mathbb Z}\alpha^\vee+{\mathbb Z}\beta^\vee$ for $T$.
In fact this allows you to determine exactly what the pseudo-Levi subgroup is in each case.
For the maximal pseudo-Levis there's an easier trick to find non-simply-connected ones: if $s\in T$ and $L=Z_G(s)$ then $Z(L)/Z(L)^\circ$ is generated by $s$, by a result of Eric Sommers. So we can see almost immediately that hardly any maximal pseudo-Levi subgroups are simply-connected. For example, the pseudo-Levi of $F_4$ which is of type $C_3\times A_1$ has a cyclic centre, so it can't be isomorphic to $Sp_6\times SL_2$. Specifically, it is isomorphic to $(Sp_6\times SL_2)/\{ \pm (I,I)\}$.
EDIT: A mistake with this is that Sommers' result only holds for adjoint type groups. More generally we have $Z(L)/(Z(L)^\circ Z(G))$ is generated by $s$. Of course this makes no difference for type $F_4$.
Does this help?
$K(X) \otimes \mathbb{Q}$ is a graded ring in the sense that it is isomorphic (by the Chern character map) to $A(X) \otimes \mathbb{Q}$, which is graded. I would perhaps prefer to say that it is "gradeable", since the grading isn't very obvious in terms of $K$-theory. The most $K$-theoretic way I know to describe it is that the Adams operators $\psi^k$ act by $k^j$ on the $j$-th graded piece. The corresponding descending filtration is the filtration by codimension.
For example, let $L$ be a line bundle with $c_1(L) = D$. Then $ch(L) = e^D$. So $ch(\log L)
=D$ and $\log [L]$, defined as the class $([L]-1) - ([L]-1)^2/2 + ([L]-1)^3/3 - \cdots$ in $K(X)$, is pure of degree $1$. Indeed, $\psi^k [L] = [L^k]$, so $\psi^k \log [L] = \log [L^k] = k \log [L]$. Let's abbreviate $\log [L]$ by $z$. The structure sheaf of the vanishing locus of a section of $L$ has $K$-class $1-L^{-1} = 1-e^z = z-z^2/2+z^3/6-z^4/24+\cdots$. So this class is in the filtered part that has degree $\geq 1$, but is not of pure degree.
A natural question, to which I don't know the answer, is whether the integer $K$-ring has a natural grading $K(X) \cong \bigoplus K_i(X)$, which turns into this grading when we tensor by $\mathbb{Q}$.
Best Answer
Since $\operatorname{Sp}_{2\ell}$ is simply-connected, there is no need for the maximal torus. So the question is about the triviality of $\operatorname{K_1}(\mathsf{C}_\ell, R) = \operatorname{Sp}(2\ell,R)/\operatorname{Ep}(2\ell,R)$, where $\operatorname{Ep}(2\ell,R)$ is the elementrary symplectic group, that is, the subgroup of $\operatorname{Sp}(2\ell,R)$ generated by its root subgroups.
A lot of cases where $\operatorname{K_1}(\mathsf{C}_\ell,R)$ is trivial are not specific fot the symplectic case and hold for all Chevalley groups. Below is a(n incomplete) list. I only consider commutative rings.