Every hypersurface is orientable? I think that every hypersurface is orientable. 
Suppose I have a hypersurface $X$ that sits inside the ambient Euclidean space $\mathbb{R}^n.$ Then for every $x \in X, T_x(X)$ is an $n - 1$ dimensional vector space. Hence, we can find a normal vector to the vector space. We can simply do a similar procedure to the boundary orientation where we define the orientation of a basis of $T_x(X), \{v_1, \ldots, v_{n - 1}\},$ as just the orientation of $\{n_x, v_1, \ldots, v_{n - 1}\},$ where $n_x$ is a normal vector to the vector space. I doubt this is correct but I fail to see where this falls apart. 
 A: As people have pointed out above in the comments, you crucially need $X \subset \Bbb R^n$ to be closed (i.e., a compact manifold without boundary) hypersurface.
In that case, consider the restriction $\underline{\Bbb R}^n = T\Bbb R^n|_X$ of the tangent bundle to the ambient Euclidean space to $X$. $TX \subset \underline{\Bbb R}^n$ is a subbundle of codimension $1$, so we take it's orthogonal complement $NX$, which is the normal bundle of the submanifold. For your procedure to be coherent, you want a smooth section of the $1$-dimensional line bundle $NX$, which would be the family of normals $\{n_x\}_{x \in X}$ you describe. 
Let $N_1 X = \{(x, v) \in NX : \|v\| = 1\}$ be the unit normal bundle; as $NX$ has rank $1$, this is a $S^0$-bundle on $X$ which is the same thing as a double cover on $X$. If this double cover wasn't trivial, there would be an embedded loop $\gamma$ in $X$ based at some point $x$ which lifts to a nontrivial embedded path $\tilde{\gamma}$ by the covering projection $N_1 X \to X$, in the sense that $\tilde{\gamma}(0) = y_0$ and $\tilde{\gamma}(1) = y_1$ where $\pi^{-1}(x) = \{y_0, y_1\}$. Let $\sigma$ be the segment in the fiber of the normal projection $NX \to X$ over $x$ joining $y_0$ and $y_1$. Then $C = \tilde{\gamma} \cup \sigma$ is a closed embedded loop in $NX$ which intersects $X$ transversely at a single point. By the tubular neighborhood theorem, we can embed $NX$ as an open neighborhood of $X$ in $\Bbb R^n$. 
This would give a $1$-dimensional submanifold $C$ and a codimension $1$ submanifold $X$ in $\Bbb R^n$ with mod $2$ intersection number $1$. But $\Bbb R^n$ is simply connected, so $C$ is nullhomotopic, which is a contradiction as that would force the mod $2$ intersection number of $C$ and $X$ to be $0$.
Therefore, $N_1 X \to X$ is the trivial cover. In particular it admits a global section $n : X \to N_1 X$. $n(x)$ is a unit normal to $X$ at $x$ by construction, so this is the desired smooth family of normals to $X$ that you wanted. Now simply orient $T_x X$ according to the orientation of $T_x \Bbb R^n$ and the normal vector $n(x)$ by the "right-hand rule", following your idea.
