See Are closed, properly embedded manifolds of co-dimension 1 in $\mathbb{R}^n$ orientable? for treatment of the smooth case.
If the topological case of Jordan theorem holds for such manifolds, then maybe one can avoid the use of Tubular Neighborhoods or orientation via normal vectors to orient them.
If $X\subset \mathbb R^n$ is a compact and locally contractible, then $H_i(X,\mathbb Z)$ is torsion free for $i=n-1,n-2$.[This is consequence of Alexander Duality](see Hatcher corollary 3.46 for details)
Now if $X$ is non-orientable closed $k$ dimensional manifold then $H_{k-1}(X,\mathbb Z)$ has a $\mathbb Z_2$ torsion. Now as you asked in your question, if it is non-orientable, then it is contradicting the first statement. So any codim 1 closed sub manifold of $\mathbb R^n$ is orientable.
Also you can see that if you exclude the closedness condition, then it is not true. For example $Mobius$-strip can be embedded in $\mathbb R^3$.
