The simple example of strongly equivalent metrics in a comment of Willie Wong actually holds the key to what is going on:
Definition. Two generalized functions $\mu_1$ and $\mu_2$ on the unit sphere $S^n \subset \mathbb{R}^{n+1}$ will be called comparable if there exists a constant $\lambda \geq 1$ such that both $$ \lambda \mu_1 - \mu_2 \; \; \hbox{and} \; \; \mu_2 - \frac{1}{\lambda}\mu_1 $$ are positive distributions (and hence measures).
Definition. A norm $\|\cdot\|$ on $\mathbb{R}^{n+1}$ is generated by a generalized function $\mu$ on $S^n$ if $$ \|v\| = \langle \mu \, | \, \xi \mapsto |\xi \cdot v| \rangle $$ for every vector $v \in \mathbb{R}^{n+1}$ (see this MO question for more on this topic).
The following simple result generalizes Willie's example:
Proposition. Comparable generalized functions generate strongly equivalent norms.
Proof. Let $\mu_1$ and $\mu_2$ be two generalized functions generating the norms $\|\cdot\|_1$ and $\|\cdot\|_2$, respectively. Asumme the generalized functions are comparable and let $\lambda \geq 1$ be such that $$ \lambda \mu_1 - \mu_2 \; \; \hbox{and} \; \; \mu_2 - \frac{1}{\lambda}\mu_1 $$ are measures. Note that, as a consequence, $$ q(v) := \langle \lambda \mu_1 - \mu_2 \, | \, \xi \mapsto |\xi \cdot v| \rangle \; \; \hbox{and} \; \; p(v) := \langle \mu_2 - \frac{1}{\lambda}\mu_1 \, | \, \xi \mapsto |\xi \cdot v| \rangle $$ are seminorms. Writing down the condition for strong equivalence we see that this immediately yields the strong equivalence of the norms. Q.E.D.
I'm guessing that if things are set up correctly, there is a converse to this proposition that completely settles the OP. Let me just add that in two dimensions we can restrict to working with measures (more complicated generalized functions are not needed to generate norms).
Corollary. Two Minkowski norms on the plane are strongly equivalent.
Proof. Every Minkowski norm $\|\cdot\|$ on the plane can be written uniquely as the cosine transform of a smooth, positive function $f : S^1 \rightarrow \mathbb{R}$ that is invariant under the antipodal map ($f(\theta) = f(-\theta)$): $$ \|(v_1,v_2)\| = \int_0^{2\pi} |\cos(\theta)v_1 + \sin(\theta)v_2|f(\theta) \, d\theta. $$ It is then clear that two Minkowski norms are generated by comparable generalized functions and are hence strongly equivalent.