A quadratic form represents all primes except for the primes 2 and 11.  By computer calculations, I found the following conjecture that the quadratic form $4x^2 + 2xy + 3y^2 + 4w^2 + 2wz + 3z^2$ represents all primes except for the two primes 2 and 11. Is it possible to prove the conjecture? Or, are there results to attack the conjecture? 
 A: As GH suggests, here the relevant Eisenstein and cusp spaces
are small enough that everything can be done explicitly.
It's even a bit better than the dimensions $5+4$ suggest,
because our quadratic form is isodual, which puts its
theta series in an eigenspace for the Atkin-Lehner involution $w_{44}$.
The resulting formula is particularly nice for $n$ prime,
and immediately shows that every prime other than $2$ and $11$
is represented, and indeed the number of representations is
proportional to the number of points modulo the prime of an
elliptic curve of conductor $11$.
Namely: let
$$
E_2(q) = 1 - 24 \sum_{n=1}^\infty \frac{nq^n}{1-q^n};
$$
this is not a modular form, but for every factor $d|44$ the combination
$$
\varepsilon^{(d)}_2(q) := d \cdot E_2(q^d) - \frac{44}{d} E_2(q^{44/d})
$$
is a weight-2 form for $\Gamma_0(44)$.  Let
$$
\phi(q) = q \prod_{n=1}^\infty \bigl( (1-q^n)(1-q^{11n}) \bigr)^2
= q - 2 q^2 - q^3 + 2 q^4 + q^5 + 2 q^6 - 2 q^7 \cdots
$$
be the unique eigen-cuspform for $\Gamma_0(11)$, associated to the
elliptic curve $E: y^2+y=x^3-x^2$ of discriminant $-11$.  Then the
theta function $\sum_{n=0}^\infty r(n) q^n$ is
$$
-\frac
  {\varepsilon^{(1)}_2(q) - \varepsilon^{(2)}_2(q) + \varepsilon^{(4)}_2(q)}
  {30}
- \frac45\bigl(\phi(q)+3\phi(q^2)+4\phi(q^4)\bigr).
$$
The coefficients are obtained by matching $q$-expansions to $O(q^{125})$,
which is more than enough to prove that two weight-$2$ forms on $\Gamma_0(44)$
coincide.  In particular, for the number of representations of a prime
$p$ other than $2$ and $11$ we have
$$
r(p) = \frac45 (p + 1 - a_p)
$$
which is positive because $p+1 - a_p$ is the number of points on $E \bmod p$
(which is indeed divisible by $5$ because $E$ has a rational $5$-torsion point
$x=y=0$).
A: There is a standard way to decide this question using modular forms or the Kloosterman refinement of the circle method (although the computational details might be tiresome).
If $r(n)$ denotes the number of representations over the integers, then $f(z)=\sum_{n=0}^\infty r(n)e(nz)$ is a modular form in $M_2(\Gamma_0(44))$, see Corollary 4.9.5 in Miyake: Modular Forms. We can write $f(z)$ as a linear combination of Eisenstein series and cuspidal Hecke eigenforms (including oldforms). Correspondingly, $r(n)$ decomposes uniquely as $r_\text{gen}(n)+r_\text{cusp}(n)$. Let us assume that $n$ is square-free and coprime with $22$. Then $r_\text{gen}(n)$ is supported on a union of arithmetic progressions mod $968$ and it is either zero or $\gg n^{1-\epsilon}$. In contrast, $r_\text{cusp}(n)\ll n^{1/2+\epsilon}$. In both estimates the implied constant depends only on $\epsilon>0$ and is effective, hence we can decide which $n$'s (square-free and coprime with $22$) are represented. I should add that $r_\text{gen}(n)=0$ implies $r(n)=0$, hence in fact  $r(n)\gg n^{1-\epsilon}$ when $r(n)>0$.
P.S. Perhaps the computational details are not so bad: the Eisenstein subspace of $M_2(\Gamma_0(44))$ has dimension $5$, while the cuspidal subspace has dimension $4$.
