Yes, I think so. Let's show that every compact set $K\subset \Omega$ is contained in some compact contractible subset of $\Omega$. We use the fact that in a simply connected open subset of the plane every compact set is contained in some compact contractible set.

Denote by $P_t$ the plane $\mathbb R^2\times t$, and define the set $\Omega_t\subset\mathbb R^2$ by $\Omega_t\times t=\Omega\cap P_t$. Define $K_t$ likewise.

For each $t$ choose a compact contractible set $C_t\subset \Omega_t$ such that $K_t\subset C_t$. There must be an interval $J_t$ containing $t$ such that for every $t'\in J_t$ we have $K_{t'}\subset C_t\subset \Omega_{t'}$.

The set of all $t$ such that $K_t$ is nonempty can be covered by finitely such intervals. Thus for some $a$ there are real numbers $s_0<\dots <s_a$ and numbers $t_i\in [s_{i-1},s_i]$ such that
$$
K\subset \cup_{i=1}^a ([s_{i-1},s_i]\times C_{t_i})\subset \Omega.
$$
Enlarge this union to make it contractible by choosing, for each $i=1,\dots a-1$, a compact contractible set $D_i$ such that $C_{t_i}\cup C_{t_{i+1}}\subset D_i\subset \Omega_{t_{i-1}}$ and then adding the sets $s_i\times D_i$.

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