This question is infuriating. I think I've made some progress, and would like to hear other's thoughts. Therefore, I am making this post a community wiki:

First of all, any triangle of area A contains a rectangle of area A/2. Proof: let the triangle be ABC, with AC the longest side. Let P and Q be the midpoints of AB and BC, and let R and S be the feet of perpendiculars from P and Q to AC. Then PQSR is a rectangle of the required area. Conversely, a rectangle of area A contains a triangle of area A/2. So we may instead ask whether there are large rectangles in the complement.

This is convenient because specifying a triangle involves 6 parameters, while specifying a rectangle has only 5. So this cuts our search space down a dimension. I find the most conveninent parameters for a rectangle to be the length of the longer side, L, the area, A, the angle of the longer side, $\theta$, and one of the vertices $(x,y)$. I'll call the type of the rectangle $(L, A, \theta)$, forgetting the translation parameters.

I now have a conjectural solution, although I have no idea how to prove that it works. I call this the sunflower configuration, because it was inspired by pattern of florets at the center of a sunflower.

Let $\tau$ be the Golden ratio. Consider the sequence of points $(\sqrt{k} \cos(2 \pi \tau k), \sqrt{k} \sin(2 \pi \tau k))$. A circle of radius $R$ around $0$ contains $R^2$ points, and has area $\pi R^2$, so the density is right.

Why do I think this is reasonable? There is a gorgeous property of the Golden ratio: if you look at the sequence $k \tau, (k+1) \tau, (k+2) \tau, ..., \ell \tau$ in $\mathbb R/\mathbb Z$, then the largest gap between any two consecutive angles is $\dfrac\tau{\ell-k}$. So multiples of $2 \pi \tau$ are very well distributed around the unit circle. I learned this from Volume 2 of the Art of Computer Programming; I'll try to find a reference later if no one else does.

I don't have a strategy yet for proving that the sunflower pattern doesn't contain large triangles (or rectangles). But, where ever I try to place one, heuristics indicate that this equidistribution of angles destroys me.

My first strategy, aiming to prove a "no", was to overlay several lattices which were all rotations of each other. Let's fix $A$ once and for all, our goal is to exclude a rectangle of size $A$. I first set out to see when a lattice could contain a rectangle of type $(L, A, \theta)$.

Translate the rectangle so that one of the short sides touches $(0,0)$. Then the rectangle contains a circular wedge of radius $L$ and angle approximately $A/L^2$ with no lattice points. In other words, there is no $(p,q)$ with $\sqrt{p^2+q^2} \le L$ and $|\theta - \tan^{-1}(p/q)| \le c\cdot A/L^2$, where $c$ is a constant I have not computed.

There are now a bunch of nuisances, having to do with the presence of that $tan^{-1}$ and the fact that people who do Diophatine approximation usually ask for $q$ to be small, not $\sqrt{p^2+q^2}$. Passing over all of the details, the set of $\theta$'s for which such a rectangle exists should look something like the union of all intervals in the $L$-th Farey sequence whose length is greater than $A/L^2$. Heuristics give me that the size of this is $c/A$ for $c$ a (different) constant that I haven't computed.

So, one lattice can't save me. The above suggests that, for every $L$, there will be a positive length set of $\theta$'s for which rectangles of type $(L, A, \theta)$ exist. Of course, we already knew that a single lattice couldn't work.

What about two lattices, rotated by some phi? I think I still lose. As $L$ grows, the set of theta's which work stays of size $1/A$, but becomes more and more spread out. Eventually, I would expect that it would contain two points that differ by phi. This part is very nonrigorous, though. In particular, I'm not sure whether we might be able to save things for some very special phi.

That's about how far I've gotten. I tried some other ideas, but didn't get anywhere. I'm not even sure which answer I think is right.