I can achieve $L(f - g) \leq \frac{1 + \pi}{2}\epsilon = (1.285\ldots)\epsilon$. Two reductions: (1) we can assume $|f(t)| < c$ for all $t \in (a,b)$ and (2) we can take $\epsilon = 1$.
(1) because $C = \{t: |f(t)| \geq c\}$ is a closed subset of $[a,b]$, so its complement is a countable set of disjoint open intervals $[a_i, b_i]$ such
that $|f(a_i)| = |f(b_i)| = c$ and $|f(t)| < c$ for all $c \in (a_i,b_i)$; then we can set $g = f$ on $C$ and handle each of these intervals separately. (2) just by scaling.

Assuming these reductions, define $g: [a,b] \to \mathbb{C}$ by letting $g(a) = f(a)$, $g(b) = f(b)$, and letting $g(t)$ move along the $|z| = c$ circle from $f(a)$ to $f(b)$ at uniform speed. The greatest possible discrepancy between $|f(b) - f(a)|$ and the length of the arc from $f(a)$ to $f(b)$ occurs when $f(a)$ and $f(b)$ are diametrically opposed and the arc length is $\frac{\pi}{2}$ times longer then the secant line. In that case, over any small interval $[t, t + \delta]$ we have $|f(t + \delta) - f(t)| < \frac{\delta}{2}$ since $L(f) < \frac{1}{2}$ by hypothesis, and $|g(t + \delta) - g(t)| \approx \frac{\pi \delta}{2}$ since small segments of a circle are approximately straight lines. Just from this we get $|(f - g)(t + \delta) - (f - g)(t)| \lessapprox \frac{\delta}{2} + \frac{\pi\delta}{2} = \frac{1 + \pi}{2}\delta$, so that $L(f - g) \leq \frac{1 + \pi}{2}$.