For $G(R) = SL_n(R)$ (see, e.g., this MO post) or $G(R) = Sp_{2n}(R)$, there is a common line of reasoning in order to prove that the reduction modulo $\mathfrak{a}$, say $\varphi_{\mathfrak{a}}: G(R) \rightarrow G(R/\mathfrak{a})$, is surjective. It consists in showing that $G(R/\mathfrak{a})$ is generated by elementary matrices any of which has a lift in $G(R)$.
For $SL_n(R)$, these elementary matrices are $E_{ij}(r) = 1_n + re_{ij}$ with $1 \le i \neq j \le n$ and where $1_n$ is the $n$-by-$n$ identity matrix and $e_{ij}$ differs from the zero matrix only by its $(i, j)$ entry, which is $r \in R$.
For $Sp_{2n}(R)$, these elementary matrices are
$
SE_{ij}(r) = \left\{
\begin{array}{ccc}
1_{2n} + re_{ij} & \text{ if } i = j',\\
1_{2n} + re_{ij} - (-1)^{i + j}re_{j'i'} & \text{otherwise}.
\end{array}\right.
$
with $1 \le i \neq j \le n$ and
where $i \mapsto i'$ is the permutation of $\mathbb{N}$ defined by $(2i)' = 2i - 1$ and $(2i - 1)' = 2i$; alternatively, this is the infinite product of transpositions $(12)(34)(56) \cdots$.
If $SL_n(R)$ is generated by the matrices $E_{ij}(r)$ with $r \in R$, the ring $R$ is said to be $GE_n$-ring in P. M. Cohn's terminology [1]. There are many examples of rings which are, or which are not $GE_n$-rings, see [1, 3] or this MO post. I don't know about the terminology for $Sp_{2n}(R)$, but here is a claim which addresses both groups.
Claim. Let $R$ be a Noetherian one-dimensional domain, e.g., $R$ is a principal ideal domain. Let $G(R) = SL_n(R)$ or $G(R) = Sp_{2n}(R)$. Then $\varphi_{\mathfrak{a}}$ is surjective for any ideal $\mathfrak{a}$ of $R$.
Proof. We can assume that $\mathfrak{a} \neq 0$. Since $R/\mathfrak{a}$ is zero-dimensional, its Bass'stable rank is $1$. Therefore $G(R/\mathfrak{a})$ is generated by elementary matrices, see [1] and [2, Theorem 7.3.b]. Because any of these matrices has a lift in $G(R)$, the map $\varphi_{\mathfrak{a}}$ is surjective.
As explained by Marty, the above should apply more generally to simply connected Chevalley groups. Let $\Phi$ be a reduced irreducible root system, $R$ be a commutative ring, and let $G = G(\Phi, R)$ be the simply connected Chevalley group of type $\Phi$ over $R$. Let $E(\Phi, R)$ be the elementary subgroup of $G$.
Claim to be checked. Let $\Phi$ be a reduced irreducible root system and let $R$ be a Noetherian one-dimensional domain. Then $\varphi_{\mathfrak{a}}$ is surjective for any ideal $\mathfrak{a}$ of $R$.
Proof. The condition on $R$ should be sufficient to ensure that
$G(\Phi, R/\mathfrak{a}) = E(\Phi, R/\mathfrak{a})$ for any non-zero ideal $\mathfrak{a}$.
[1] P. Cohn, "On the structure of the $GL_2$ of a ring", 1966.
[2] L. Vaserstein, A. Suslin, "Serre's problem on projective modules over polynomial rings, and algebraic $K$-theory", 1976.
[3] F. Grunewald, J. Mennicke and L. Vaserstein, "On the groups $SL_2(\mathbb{Z}[x])$ and $SL_2(k[x,y])$", 1994.
