The answers to both your questions are yes. (ie for any trace class operator $a$ on a Hilbert space $H$ with $Tr(a)=0$, there exists an orthonormal basis of $H$ consisting of vectors $\xi$ such that $\langle a \xi,\xi\rangle=0$).
First note that the fact that there exists an orthonormal basis of vectors $\xi$ such that $\langle a \xi,\xi\rangle=0$ is immediate (by something like "transfinite induction") from the fact that for any $a$ with trace zero, there exists one non-zero vector $\xi$ with $\langle a \xi,\xi\rangle=0$. Indeed, Zorn's Lemma implies that there exists a maximal closed subspace $H$ of $K$ such that there is an orthonormal basis of $K$ consisting of vectors such that $\langle a \xi,\xi\rangle=0$. If $K$ were different from $H$, consider $P$ the orthonormal projection on the orthogonal of $K$. The restriction of $Pa$ to the orthogonal of $K$ is still of trace class and zero trace, so there exists a unit vector in the orthogonal of $K$ such that $\langle a \xi,\xi\rangle=0$, which contradicts the maximality of $K$.
Now to the first point. You want to prove that is $Tr(a)=0$, zero belongs to the numerical range $W(a)$ of $a$, which is the set of $\langle a \xi,\xi\rangle$ for $\xi$ in the unit sphere of $H$. $W(a)$ is a convex subset of $\mathbb C$ for any operator $a$. This is usually stated for matrices (see this page), but since everything happens in two dimensions this remains true in general.
But the assumption $Tr(a)=0$ implies that there is a sequence $(\lambda_n)_n$ in $W(a)$ such that $\sum_n \lambda_n=0$. This clearly implies that $0$ belongs to the convex hull of the $\lambda_n$'s, which is contained in $W(a)$.
Edit: To answer Bill's objection, here is a proof that if $\sum_n \lambda_n=0$, then $0$ belongs to the convex hull of the $\lambda_n$'s. (I agree that for a general compact operator, $0$ does not always belong to $W(a)$ but only to its closure, but my point was that here the assumption $Tr(a)=0$ makes it different).
Assume, by contradiction, that $0$ does not belong to the convex hull of the $\lambda_n$'s. By (some form of) Hahn-Banach, the $\lamdba_n$'s \lambda_n$'s all belong to some closed half-plain, say for simplicity $Im(\lambda_n)\geq 0$ for all $n$. The equality $\sum_n Im(\lambda_n)=0$ then implies that the $\lambda_n$'s are all real. They cannot be all positive (or all negative), otherwise $\sum_n \lambda_n \ne 0$. This is a contradiction.