This answer builds on Bill Johnson's comment.
This is not a full answer since a step is missing, but too long for a comment.

A Lipschitz function is almost everywhere differentiable, so let $a\in Q$ be a point where $f$ is differentiable.
For simplicity, we can take $a$ to be an interior point.
There is a $2\times2$ matrix $A$ and a continuous function $R:Q\to\mathbb R^2$ so that
$$
f(a+x)=f(a)+Ax+|x|R(x)
$$
and $R(0)=0$.

(This paragraph is not necessary, but may give better insight to the next one.)
If $A=0$, then $|f(a+x)-f(a)|\leq|x||R(x)|$ and so $\text{diam}(f(B(a,r)))\leq2r\sup_{B(0,r)}|R|$.
Since the width of $f(B(a,r))$ is at least $cr$, its diameter also has this lower bound.
Thus $c\leq2\sup_{B(0,r)}|R|$ for all $r>0$.
But this is impossible since the supremum tends to zero as $r\to0$.
Therefore $A\neq0$.

Suppose $A$ does not have full rank and let $v$ be a unit vector perpendicular to its image.
Now
$$
v\cdot(f(x+a)-f(a))
=
|x|v\cdot R(x).
$$
The width of $f(B(a,r))$ is bounded from above by its width in the direction of $v$, which in turn is bounded by
$$
\sup_{y\in B(a,r)}|v\cdot(f(y)-f(a))|
=
\sup_{x\in B(0,r)}|x||R(x)|.
$$
As this is bounded from below by $cr$, we get
$$
c
\leq
\sup_{x\in B(0,r)}|R(x)|
$$
for all $r>0$.
This is a contradiction since the supremum tends to zero as $r\to0$ by continuity of $R$.
Thus $A$ has full rank.

The mapping $x\mapsto f(x)-Ax$ is Lipschitz.
Suppose one can show that the Lipschitz constant can be made arbitrarily small by restricting it to $B(a,r)$ for $r$ small enough.
(This would be easy if $f$ was continuously differentiable.)
If this Lipschitz constant is less than $\frac12\|A^{-1}\|^{-1}$, then $f(x)=f(y)$ implies
$$
|x-y|
=
|A^{-1}(f(x)-Ax-f(y)+Ay)|
\leq
\frac12|x-y|
$$
and thus $x=y$.
Thus $f$ is injective in a small neighborhood of $a$.
Then it follows from invariance of domain that $f(a)$ is an interior point of $f(Q)$.