Let me try to reinstate honor to the solution that proposed the basis change, by **reducing** the general case to the independent ("generic") case via a basis change, as opposed to proving it from scratch. This time I am treating the **continuous version**. (as in Peter Humphries' solution)

Let $S$ be the subspace of vectors $(x_1,\ldots,x_k)$ such that
$$a_1x_1+\dots+a_kx_k=0$$
for all rational vectors $(a_1,\ldots,a_k)$ for which
$$a_1\theta_1+\dots+a_k\theta_k=0.$$
The integer points in $S$ form a [lattice](https://en.wikipedia.org/wiki/Lattice_(group)) $L$ (a discrete subgroup of $\mathbb R^n$), 
which can be written as $L=\{\,c_1b_1+\dots+c_rb_r\mid c_i\in\mathbb Z\,\}$ for some
generating
basis vectors $b_1,\dots,b_r\in\mathbb{Z}^k$. Since $S$ is defined by rational equations, this basis spans $S$. Let $(\theta'_1,\dots,\theta_r')$ be the coordinates of the point $(\theta_1,\dots,\theta_k)\in S$ with respect to this basis.
Then $(\theta'_1,\dots,\theta_r')$ is independent over the rationals (see below (*) for a proof). Thus, by the generic continuous Kronecker-Weyl Theorem,
$(t(\theta_1',\dots,\theta_r'))_{t\in \mathbb{R}}$ is uniformly distributed modulo 1 in the $r$-torus $[0,1)^r$.
Transforming back, $(t(\theta_1,\dots,\theta_k))$ is uniformly distributed modulo $L$ in the fundamental region
$$ F = \{\,\lambda_1b_1+\dots+\lambda_rb_r \mid
0\le\lambda_i<1\,\}$$
of the lattice. Now we map $F$ back into the standard torus $[0,1)^k$ by taking all coordinates modulo 1. No two points of $F$ are mapped to the same point (otherwise we would have an integer point inside $F$), but
"opposite" boundary points _are_ mapped to the same point because they differ by a basis vector $b_i$. So $F$ forms a nice $r$-dimensional subtorus of $[0,1)^k$.
<hr>
(*) Here is the proof. It is not so obvious as I thought. Suppose there is a nontrivial rational relation $c_1\theta_1'+\dots+c_r\theta'_r=0.$ We can choose inside $S$ a rational vector $a$ such that $\langle a_i,\theta'_i\rangle = c_i$ for $i=1,\dots,r$. (The vector $a$ is uniquely determined by these equations.) Then
$$\langle a, (\theta_1,\dots,\theta_k)\rangle= \langle a, (\theta_1'b_1+\dots+\theta_r'b_r)\rangle =c_1\theta_1'+\dots+a_r\theta'_r=0
$$
Thus, by the definition of $S$, $a$ should be orthogonal to $S$. This is a contradiction.