My answer is **Yes**. Of course, I presume that you assume $\Omega$ is on one side only of its boundary.

The boundary $\partial\Omega$ is compact. By assumptions, it has an atlas with finitely many charts, each one corresponding to a piece $\Gamma_j$ which is the graph of a Lipschitz function: in appropriate coordinates, $\Gamma_j$ is given by $x_d=\phi_j(x_1,\ldots,x_{d-1})$.

Let us choose $\delta$ smaller than one tenth of
$$\delta_0:=\min_{x\in\bar\Omega}\\,\max_j\{d(x,\partial\Gamma_j)\}>0$$

Let $\bar x$ be on the boundary of $\Omega_\delta$, not on $\partial\Omega$. Let $j$ be such that $B(\bar x;9\delta)\cap\partial\Omega\subset\Gamma_j$. There exists a point $\bar y\in\partial\Gamma_j$ such that $d(\bar x,\bar y)=\delta$. Wlog, we have $\bar y=0$ and the graph is locally $x_d=\phi_j(x_1,\ldots,x_{d-1})$. Let $M$ be the Lipschitz constant of $\phi_j$. Locally, the boundary of $\Omega_\delta$ is the upper envellop of the balls $B(y;\delta)$ with $y\in \Gamma_j$; as a matter of fact, for $x\in B(\bar x;\delta)$, the closest points of $\partial\Omega$ to $x$ belong to $\Gamma_j$. It is easy to see that it is the graph of a function $\phi_j^\delta$ whose Lipschitz constant is at most $M$.