The definition of "plumbed manifold" that I'm using in this context is the following - given a weighted tree $\Gamma$, build up a framed link $L(\Gamma)$ by chaining together two copies of the unknot whenever two vertices of $\Gamma$ are adjacent. Then the plumbed manifold $Y(\Gamma)$ is the result of performing integral Dehn surgery on $L(\Gamma) \subseteq S^3$ with surgery data specified by the framing of each component.
It's known that any closed orientable $3$-manifold can be obtained by integral Dehn surgery on a framed link $\mathcal{L}$ where every component of $\mathcal{L}$ can be taken to have even framing (i.e. the integer which specifies the framing is even). This is a slightly enhanced version of the Lickorish-Wallace theorem which states that any closed orientable $3$-manifold can be obtained by integral Dehn surgery on a framed link $\mathcal{L}$.
My question is - can every plumbed manifold arise from a framed link $L(\Gamma)$ built up from a weighted tree $\Gamma$ (as specified above) such that $L(\Gamma)$ has components that have only even framing.
To perhaps phrase this more conretely, can every plumbed manifold arise from a framed link $L(\Gamma)$ which consists of only unknotted components with even framings and is such that $L(\Gamma)$ contains no "loops" of the components, i.e. if we write $K_iK_j$ to mean the $i$ith an $j$th unknotted components of $L(\Gamma)$ are chained together, then no sequence of the form $K_1K_2\dots K_nK_1$ exists.
The problem one runs into when trying to apply the proof for the "enhanced version" of the Lickorish-Wallace theorem is that the Kirby moves, which allow one to modify the framing numbers such that they are all even, might induce non-trivial knotted components in the link $L(\Gamma)$ and the resulting manifold will not be a plumbed manifold.