Unless I am terribly mistaken, the answer is yes and the proof strategy is rather simple. The crucial observation is that the definition of bounded geometry depends on quantities that are continuous.

Consider first the injectivity radius function of the boundary, $r_{b}\colon\delta X\to \mathbb{R}$,
$$r_b(x)=\sup\{\ t>0\mid \exp\colon B_{\delta X}(0_x,t)\to \delta X \quad \text{is a diffeomorphism}\ \}.$$ This is a continuous and positive function on $\delta X$ and it is bounded away from zero on $\delta X\setminus K$, so it must be bounded away from zero over all $\delta X$ because $K$ is compact. Denote the positive lower bound by $r_b(X)$; this is the injectivity radius of $X$.

Now define the ''normal collar injectivity radius'' $r_C\colon \delta X\to\mathbb{R}$ as $$r_c(x)=\sup\{\ t>0\mid \kappa\colon B_{\delta X}(x,r_b(X))\times[0,t)\to X\quad \text{is well-defined and a diffeomorphism} \ \},$$
where $\kappa(x,t)=\exp(t\nu_x)$ and $\nu_x$ is the unit inward normal vector. Again, this is continuous and bounded away from zero outside a compact set, hence bounded away from zero; we have obtained that $X$ satisfies the normal collar condition in Schick's definition.

A similar argument works for the remaining two conditions.

**Answer to the added question**: If $E$ is a manifold with boundary of bounded geometry with the metric that is the restriction of that of $\overline{E}$, then $E=\overline{E}$.

Assume for the sake of contradiction that $E$ is of bounded geometry and there is a point $x\in \overline{E}\setminus E$. Consider first the case where $x$ is in the closure of $\delta E$, so $x\in \delta \overline{E}$ and there is a sequence $(y_n)$ in $\delta E$ with $y_n\to x$. The injectivity radius $r_b$ of $\delta E$ is positive, so $B_{\delta \overline{E}}(y_n,r_b)\subset \delta E$ and $d_{\delta \overline{E}}(x,y_n)>r_b$ for all $n$, a contradiction.

Similar arguments apply when $x$ is a limit of points in the normal collar or a limit of points not in the normal collar.