Are most cubic plane curves over the rationals elliptic? %This is a new version of the original question modified in the light of the answers and comments.
The word 'most' in the title is ambiguous. The following is one way of making it precise.
Question1: (This seems to be open. See Poonen's answer below)
A cubic projective curve over $\mathbb{Q}$ is given by ten relatively prime  integers (the coefficients of its equation after clearing the denominators). Suppose we take a ten dimensional box $[-N,N]^{10}$ and choose points with integer coordinates with respect to the uniform measure and form the equation of the associated cubic curve. Suppose the number of points which give rise to a curve with a rational point is $E(N)$. Then what can we say about $E(N)/(2N+1)^{10}$ as $N\rightarrow \infty$?
Should the limit exist and if it does, should it be one, zero, or some other number?
Another question of interest is:
Question 2: (There is a satisfactory answer to this. See Voloch's response below.)
Are either of the sets {cubics with no rational point} and {cubics with at least one rational point} Zariski dense?
 A: Re: Zariski denseness. Let $Y$ be the set of cubics with a rational point and $N$ the set of cubics without a rational point, both viewed as subsets of $P^9$. I claim that both $Y$ and $N$ are Zariski dense. For $Y$, note that if you fix a point the plane, the set of cubics containing that point is a hyperplane in $P^9$ so the Zariski closure of $Y$ contains infinitely many hyperplanes and therefore is the whole of $P^9$. For $N$, we use Bjorn's example. Any curve congruent modulo 8 to $x^3+2y^3+4z^3=0$ is in $N$ and this set of curves is already Zariski dense.
A: "One could try to estimate the size of the Tate-Shafarevich group of a "random" elliptic curve, to get an idea of how often local solvability implies global solvability, but even if one does this it is not clear that this is counting curves in the same way."
Bhargava has reportedly proven the 3-Selmer group has average size 4. The assumption of a minimal rank (1/2 average) and Parity conjecture would account that 2 of the 4 come from rank, and 2 of the 4 come from Sha, so 50%. His counting is by $|c_4| < X^2 $ and $|c_6| < X^3$ I think. The workers on 3-descent have some bounds that relate the invariants to the coefficient size.
A: Your question (as explained in the second paragraph) is not vague at all!  In fact, it appears for instance after Conjecture 2.2 in http://www-math.mit.edu/~poonen/papers/random.pdf , which is Random diophantine equations, B. Poonen and J. F. Voloch, pp. 175–184 in: Arithmetic of higher-dimensional algebraic varieties, B. Poonen and Yu. Tschinkel (eds.), Progress in Math. 226 (2004), Birkhäuser.
The answer is not known, and the experts I've spoken to do not even have a convincing heuristic predicting an answer.  Swinnerton-Dyer told me that he had a hunch that the answer was 0, and this is my hunch too, but we have little to back this up.
It is not even clear that the limit exists.  One can prove, however, that the density (in your precise sense) of plane cubic curves that have points over $\mathbb{Q}_p$ for all $p \le \infty$ is a number strictly between $0$ and $1$ (Theorem 3.6 in the Poonen-Voloch paper), so the lim sup of the fraction of plane cubic curves with a rational point is at most this; in particular, it's not 1.
One could try to estimate the size of the Tate-Shafarevich group of a "random" elliptic curve, to get an idea of how often local solvability implies global solvability, but even if one does this it is not clear that this is counting curves in the same way.
A: Manjul Bhargava has answered this question yesterday : 
A positive proportion of plane cubics fail the Hasse principle
Manjul Bhargava
(Submitted on 5 Feb 2014)
When all ternary cubic forms over $\mathbf{Z}$ are ordered by the heights of their coefficients, we show that a positive proportion of them fail the Hasse principle, i.e., they have a zero over every completion of $\mathbf{Q}$ but no zero over $\mathbf{Q}$. We also show that a positive proportion of all ternary cubic forms over $\mathbf{Z}$ nontrivially satisfy the Hasse principle, i.e., they possess a zero over every completion of $\mathbf{Q}$ and also possess a zero over $\mathbf{Q}$. Analogous results are proven for other genus one models, namely, for equations of the form $z^2=f(x,y)$ where $f$ is a binary quartic form over $\mathbf{Z}$, and for intersections of pairs of quadrics in $\mathbf{P}_3$.﻿
