Accurate quantum state estimation via "Keeping the experimentalist honest" Bob has a black-box, with the label "V-Wade", which he has been promised prepares a qubit which he would like to know the state of.  He asks Alice, who happens also to be an experimental physicist, to determine the state of his qubit.  Alice reports $\sigma$ but Bob would like to know her honest opinion $\rho$ for the state of the qubit.  To ensure her honesty, Bob performs a measurement $\{E_i\}$ and will pay Alice $R(q_i)$ if he obtains outcome $E_i$, where $q_i={\rm Tr}(\sigma E_i)$.  Denote the honest probabilities of Alice by $p_i={\rm Tr}(\rho E_i)$.  Then her honesty is ensured if
$$
\sum_i p_i R(p_i)\leq\sum_i p_i R(q_i).
$$
The Aczel-Pfanzagl theorem holds and thus $R(p)=C\log p + B$.  Thus, Alice's expected loss is (up to a constant $C$)
$$
\sum_i p_i(\log{p_i}-\log{q_i})=\sum_i {\rm Tr}(\rho E_i)\log\left[\frac{{\rm Tr}(\rho E_i)}{{\rm Tr}(\sigma E_i)}\right].
$$
Blume-Kohout and Hayden showed that if Bob performs a measurement in the diagonal basis of Alice's reported state, then her expected loss is the quantum relative entropy
$$
D(\rho\|\sigma)={\rm Tr}(\rho\log\rho)-{\rm Tr}(\rho\log\sigma).
$$
Clearly in this example Alice is constrained to be honest since the minimum of $D(\rho\|\sigma)$ is uniquely obtained at $\sigma$.  This is not true for any measurement Bob can make (take the trivial measurement for example).  So, we naturally have the question: which measurements can Bob make to ensure Alice's honesty?  That is, which measurement schemes are characterized by
$$
\sum_i {\rm Tr}(\rho E_i)\log\left[\frac{{\rm Tr}(\rho E_i)}{{\rm Tr}(\sigma E_i)}\right]=0\Leftrightarrow \sigma=\rho?
$$
Note that $\{E_i\}$ can depend on $\sigma$ (what Alice reports) but not $\rho$ (her true beliefs).
Partial answer: Projective measurement in the eigenbasis of $\sigma$ $\Rightarrow$ yes, Blume-Kohout/Hayden showed that this is the unique scheme constraining Alice to be honest for a projective measurement.
Informationally complete $\Rightarrow$ yes, clearly this constrains Alice to be honest since the measurement will uniquely specify the state (moreover, the measurement can be chosen independent of $\sigma$).  
Trivial measurement $\Rightarrow$ no, Alice can say whatever she wants without impunity.
 A: Trivially for any set of measurements $\{E_i\}$ where $\rho$ and $\sigma$ have equal expectation value for each $E_i$, $$\sum_i\mbox{Tr}(\rho E_i) \log \left[ \frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)}\right] = \sum_i\mbox{Tr}(\rho E_i) \times 0 = 0.$$
Note that the log-sum inequality theorem says that $$\sum_i a_i \log\left(\frac{a_i}{b_i}\right) \leq \big(\sum_i a_i\big)\times \log \left(\big(\sum_i a_i\big)/\big(\sum_i b_i\big)\right),$$ with equality only if $\frac{a_i}{b_i}$ is constant for all $i$. This implies that
$$\sum_i\mbox{Tr}(\rho E_i) \log \left[ \frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)}\right] = 0$$ if and only if $\frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)}$ is constant. Since the sum of the traces over $i$ is 1 for both density matrices this implies that you must have $\frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)} = 1$.
If, however, the projections onto each subspace selected by $E_i$ aren't equal, then there exists some $i$ such that $\frac{\mbox{Tr}(\rho E_i)}{\mbox{Tr}(\sigma E_i)} \neq 1$.
Hence for Alice to have zero expected loss, then $\mbox{Tr}(\rho E_i) = \mbox{Tr}(\sigma E_i) ~~~\forall i$. If expectation value for each $E_i$ uniquely identifies a density matrix, then necessarily $\rho = \sigma$ if Alice has zero expected loss. A sufficient condition for this is when $E_i$ project onto $(\dim \mathcal{H})^2 -1$ linearly independent density matrices, where $\mathcal{H}$ is the Hilbert space of the system.
