Let $(M, \xi)$ be a contact manifold, and $\Lambda \subset M$ a closed connected Legendrian with a tubular open neighborhood $W$. The question is that, can one find a smaller tubular neighborhood $U \subset W$ of $\Lambda$, such that the boundary $\partial U$ is a smooth convex hypersurface in $M$? (A hypersurface in a contact manifold is convex if there exists a contact vector field $X$ transverse to it.)
A singular Example: let $(M = \mathbb{R}^3, \xi = \ker(\alpha = dz - y dx))$, and let the singular Legendrian $\Lambda=\Lambda_1 \cup \Lambda_2$, where $$ \Lambda_1 = \{(t,0,0)\mid t \in \mathbb{R} \}, \quad \Lambda_2 = \{(t,t,t^2/2)\mid t < 0\}. $$ (I am not sure in general what condition to impose on the singularity of $\Lambda$, maybe we assume it has a Whitney stratification $\Lambda = \sqcup S_\alpha$. In the above example, there would be 3 one-dimensional strata and 1 zero-dimensional stratum.)
The smooth case: In the case where $\Lambda$ is a smooth Legendrian, maybe one can use the following construction: (1) identify a neighborhood of $\Lambda \subset M$ with a neighborhood of $\Lambda \subset J^1(\Lambda) \cong T^* \Lambda \times \mathbb{R}_z$, with $\alpha = dz - \theta$ where $\theta$ is the Liouville form "$y dx$" on $T^* \Lambda$. (2) Then the vector field $X = z \partial_z + V$, where $V$ is the outgoing Liouville vector field "$y \partial_y$" on $T^* \Lambda$, is a contact vector field: $\mathcal{L}_{X} \alpha = \alpha$. (3) Fix an inner product $\| \|$on $T^* \Lambda$, and a small enough positive constant $r$, then the neighborhood of $\Lambda$ can be taken as $U = \{(p,z) \in T^* \Lambda \times \mathbb{R} \mid z^2 + \|p\|^2 < r\} \subset J^1(\Lambda)$.
