The space of orthonormal $k$-frames in $\mathbb{R}^n$ is the Stiefel manifold $V(k, n) = SO(n)/SO(n - k)$. There is a natural $SO(k)$ action on $V(k, n)$ and the quotient is the oriented grassmannian $\operatorname{Gr}^+(k, n) = SO(n)/(SO(k)\times SO(n-k))$. Let $\gamma_k \to \operatorname{Gr}^+(k, n)$ denote the tautological bundle and let $\gamma_k^{\perp} \to \operatorname{Gr}^+(k, n)$ denote its orthogonal complement. As you indicated, a nowhere-zero section of $\gamma_k^{\perp} \to \operatorname{Gr}^+(k, n)$ would give rise to a map that you desire. In fact, such a map arises this way if and only if it is $SO(k)$-invariant.

The inner product on $\mathbb{R}^n$ allows us to define the map $P \mapsto P^{\perp}$ which induces a diffeomorphism $f : \operatorname{Gr}^+(k, n) \to \operatorname{Gr}^+(n-k, n)$. Under this diffeomorphism we have $f^*\gamma_{n-k} \cong \gamma_k^{\perp}$, so $\gamma_k^{\perp} \to \operatorname{Gr}^+(k, n)$ admits a nowhere-zero section if and only if $\gamma_{n-k} \to \operatorname{Gr}^+(n-k, n)$ does. Therefore, we would like to know the answer to the following question:

For which values of $k$ and $n$ does $\gamma_{n-k} \to \operatorname{Gr}^+(n-k, n)$ admit a nowhere-zero section?

One necessary condition is that $w_{n-k}(\gamma_{n-k}) = 0$. Said another way, if $w_{n-k}(\gamma_{n-k}) \neq 0$, then there is no $SO(k)$-invariant map.

In a previous version of this answer, I stated what I thought was the $\mathbb{Z}_2$ cohomology ring of $\operatorname{Gr}^+(k, n)$ - I was incorrect. From this mistake, it followed that for $1 < k < n - 1$, $w_{n-k}(\gamma_{n-k}) \neq 0$ and hence there were no $SO(k)$-invariant maps for these values of $k$. This conclusion is false; there is a counterexample when $k = 2$ and $n = 7$ as David E Speyer pointed out in the comments below.

Somewhat surprisingly, the $\mathbb{Z}_2$ cohomology ring of $\operatorname{Gr}^+(k, n)$ is not known in general, see this question. The values of $k$ and $n$ for which $w_{n-k}(\gamma_{n-k}) \neq 0$ also seems to be unknown.

When $k = n - 1$, you described such a map which is in fact $SO(n-1)$-invariant. By the above correspondence, such maps exist because $\gamma_1 \to \operatorname{Gr}^+(1, n) = S^{n-1}$ is trivial as it is an orientable line bundle (alternatively, $\gamma_1$ is trivialised by the Euler vector field).

When $k = 1$, first note that $\gamma_{n-1} \to \operatorname{Gr}^+(n - 1, n) = S^{n-1}$ is isomorphic to the tangent bundle of $S^{n-1}$:

\begin{align*}
TS^{n-1} &\cong T\operatorname{Gr}^+(n-1, n)\\
&\cong \operatorname{Hom}(\gamma_{n-1}, \gamma_{n-1}^{\perp})\\
&\cong \gamma_{n-1}^*\otimes\gamma_{n-1}^{\perp}\\
&\cong \gamma_{n-1}\otimes f^*\gamma_1\\
&\cong \gamma_{n-1}
\end{align*}

where the last isomorphism uses the fact that $\gamma_1$, and hence $f^*\gamma_1$, is trivial. By Poincaré-Hopf, $TS^{n-1}$ admits a section if and only if $n$ is even. In this case, the map can be written down explicitly: $(v_1, v_2, \dots, v_{n-1}, v_n) \mapsto (-v_2, v_1, \dots, -v_n, v_{n-1})$. Identifying $\mathbb{R}^n$ and $\mathbb{C}^{n/2}$ via $(v_1, v_2, \dots, v_{n-1}, v_n) \mapsto (v_1 + iv_2, \dots, v_{n-1} + iv_n)$, the aforementioned map is nothing but multiplication by $i$.

Note, requiring $SO(k)$-invariance for $k = 1$ is not a restriction as $SO(1)$ is the trivial group.

thinkthat the obstruction you want is the Euler class, on the Wikipedia page it says "Note that "Normalization" is a distinguishing feature of the Euler class, so that it detects the existence of a non-vanishing section". So you'd need to look at the pull-back of the Euler class under the map $Gr_{n,k} \to Gr_{n,n-k}$ which maps $E$ to its orthogonal complement. The issue is a little more complicated if you're starting with a basis as then you need to pull back further to the frame bundle. – Loop Space Nov 5 at 17:41