Solving a linear diophantine equation approximately
All right, having thought about it more I believe I have an explicit reduction from this problem to extended GCD; I'll explain it in the $n=2$ case, but I believe that it extends to arbitrary $n$.  Note that this finds an $\mathbf{x}$ that minimizes the distance to the hyperplane, but not necessarily the smallest $\mathbf{x}$ (there are in fact infinitely many values that attain the same minimum distance) - I believe the latter problem is also feasible, but haven't given it any real thought yet.  The algorithm is based on a few simple principles:


If $g=\mathrm{GCD}(a_1, a_2)$, then the set of values taken on by $\mathbf{a}\cdot\mathbf{x} = a_1 x_1 + a_2 x_2$ is precisely $\left\{0, \pm g, \pm 2g, \pm 3g, \ldots\right\}$; furthermore, values $x_1$ and $x_2$ with $a_1 x_1 + a_2 x_2 = g$ can be found efficiently (this is exactly the extended Euclidean algorithm).
The minimum distance from the hyperplane to a point on the lattice is the minimal distance from a point on the lattice to the hyperplane (obvious, but a useful inversion of the problem).
The distance from a given point $\mathbf{x}$ to the hyperplane $\mathbf{a}\cdot\mathbf{y} = b$ is proportional to $\left|\mathbf{a}\cdot\mathbf{x}-b\right|$ (specifically, it's $1/|\mathbf{a}|$ times this value - but since multiplying all distances by this value has no effect on the location of the minimum, we can ignore the normalizing factor).


This suggests the following procedure:


Compute $g=\mathrm{GCD}(a_1,a_2)$, along with $x^0_1, x^0_2$ such that $a_1x^0_1+a_2x^0_2 = g$.
Compute $r = \left\lfloor{b\over g}\right\rfloor$ and calculate $d =\min(b-rg, (r+1)g-b)$; $d$ is the (scaled) minimal distance from the lattice to the hyperplane.  Let $s$ be either $r$ or $r+1$ ($=\left\lceil{b\over g}\right\rceil$, unless $b$ is a multiple of $g$) depending on which one attains the minimal distance.
Compute $x_1 = sx^0_1$ and $x_2 = s x^0_2$; then $\mathbf{a}\cdot\mathbf{x} = sg$ is the nearest multiple of $g$ to $b$, and thus $\left|\mathbf{a}\cdot\mathbf{x}-b\right|$ attains the minimum of this distance over all lattice points.


As far as I know, the exact same procedure should work correctly in arbitrary dimensions; the key is that the $n$-dimensional GCD still satisfies Bezout's identity, and so to find the minimum distance to a lattice point you only have to find the nearest multiple of $g$ to $b$.