Is it possible that two measurements will determine if the labels are right? Suppose we have $6$ identically looking cubes that weigh  $1,\dots,6$ grams, respectively. Someone has labeled them 1g, 2g, 3g, 4g, 5g,6g however he might have labeled them incorrectly.
Suppose you have a scale and you can't feel the weight by your hand, is it possible to use the scale twice to tell if the labels are correct?
Edit
The scale can only tell you if two side are equal, or that which side is heavier. The answer by @Evan means that we are allow to measure the weight of the set of cubes. Sorry for the confusion I've made.
 A: *

*Compare (6) and (1,2,3). If scales are not equal we know some labels are wrong. If they are equal, we know 6 is 6, set {1,2,3} really contains cubes with weights {1,2,3} (may be wrong labelled), set {4,5} really contains cubes weighted 4 and 5.

*Compare (6, 1) and (5, 3). This time we expect right side to be heavier. !!!
Left side weights at least 7 (because we know real 6 in on the left). And the right side can weight 8 only if it has heaviest from sets {1,2,3} and {4, 5}, that is 5 and 3.
A: THIS ANSWER WAS WRITTEN UNDER THE IMPRESSION THAT OP MEANT A SCALE LIKE THIS:

RATHER THAN A SCALE LIKE THIS:
. AFTER THE ANSWER WAS WRITTEN, OP ADDED THE CLARIFICATION THAT IT WAS MEANT THE OTHER WAY ROUND.

No, it is not possible to determine whether the labels are correct.
Contrary to what might appear at first sight, the proof is short; it is contained in the last paragraph preceding the concluding section, Generalizations. The rest of this post is dedicated to formalizing what needs be proved (the resulting formalization being the highlighted sentence near the end of this post), to introducing notation used in the formalization and in the proof, and to arguing that the formalization agrees with the problem that is described informally in the Original Post.
Denote the set of labels by $L$, i.e. $L := \{1, 2, \dots, 6\}$. Denote by $\Pi$ the set consisting of all the ways to label the cubes. For every $\pi \in \Pi$, and for every $\ell \in L$, $\pi(\ell)$ is the weight of the cube labeled $\ell$ as per the labeling schema $\pi$. In other words, $\Pi$ is the set of permutations of $L$. Denote the identity permutation by $I$, and denote the given labeling schema by $\pi^*$.
For every labeling schema $\pi \in \Pi$, denote by $w_\pi$ the function that assigns to every subset, $S \subseteq L$, of cubes, identified by their labels, their weight per $\pi$:
$$
w_\pi(S) := \sum_{s \in S} \pi(s).
$$
Every subset of labels, $S \subseteq L$, induces an equivalence relation, $\cong_S$, on $\Pi$ as follows. For every $\pi_1, \pi_2 \in \Pi$, $\pi_1 \cong_S \pi_2$ iff $w_{\pi_1}(S) = w_{\pi_2}(S)$. (Verify that this is indeed an equivalence relation!) Denote by $E_S$ the family of equivalence classes corresponding to $\cong_S$. For every labeling schema, $\pi \in \Pi$, $\pi$'s equivalence class in $E_S$ shall be denoted by $[\pi]_S$.
With this terminology, we will show that the problem reduces to the following question. Is there some pair $S_1, S_2 \subseteq L$, such that
$$
[I]_{S_1} \cap [I]_{S_2} = \{I\}\tag{1}\label{Apple}
$$
To see that this formulation indeed captures the problem statement described in the Original Post, suppose firstly that there are some $S_1,S_2 \subseteq L$ such that \eqref{Apple} holds. Use the scale to weigh the cubes, whose label set $S_1$ is, and then use the scale once more to weigh the cubes, whose label set $S_2$ is. Denote the two weights thus obtained by $w_1, w_2$, respectively. If
$$
w_1 = w_I(S_1)\ \wedge\ w_2 = w_I(S_2), \tag{2}\label{Banana}
$$
report that $\pi^* = I$; otherwise, report that $\pi^* \neq I$. To verify the correctness of this algorithm, it suffices to observe that condition \eqref{Banana} holds iff $\pi^* \in [I]_{S_1}\cap[I]_{S_2}$.
Conversely, suppose there is some pair $S_1,S_2\subseteq L$ of sets of cubes (given by their labels as per the labeling schema $\pi^*$), for which it is possible, by consecutively weighing $S_1$ and $S_2$, to tell whether $\pi^* = I$ or $\pi^* \neq I$. Any algorithm, $\mathfrak{A} = \mathfrak{A}_{S_1, S_2}$, that can be used to reach this determination, can have only indirect access to $\pi^*$ in the form of $S_1$'s and $S_2$'s weights. For every $\pi\in\Pi$, $\mathfrak{A}$ takes $w_\pi(S_1)$, and $w_\pi(S_2)$ as input arguments, and it is based solely on these arguments (as well as on $S_1$ and $S_2$) that $\mathfrak{A}$ must calculate its output: 
$$
\mathfrak{A}\big(w_\pi(S_1), w_\pi(S_2)\big) = \begin{cases}
\mathrm{TRUE} &, \pi = I, \\
\mathrm{FALSE} &, \mathrm{otherwise}.
\end{cases}
$$
To see that \eqref{Apple} holds, let $\pi \in [I]_{S_1}\cap[I]_{S_2}$. Then $\mathfrak{A}\big(w_\pi(S_1), w_\pi(S_2)\big) = \mathfrak{A}\big(w_I(S_1), w_I(S_2)\big) = \mathrm{TRUE}$, so that $\pi = I$.
For every distinct pair of labels, $a, b \in L$, denote by $\tau_{\{a,b\}}$ the transposition $(a\ b)$, i.e. $\tau_{\{a, b\}}$ is the permutation $\pi \in \Pi$ that satisfies
$$
\begin{align}
\pi(a) &= b, \\
\pi(b) &= a,
\end{align}
$$
and $\pi(\ell) = \ell$ for every $\ell \in L\setminus\{a,b\}$.
We can now finally show that there are no $S_1, S_2 \subseteq L$, such that \eqref{Apple} holds. Equivalently, we will show the following.

The set $T$, consisting of all the pairs $(S_1, S_2) \in (\mathcal{P}L)^2$ for which \eqref{Apple} holds, is empty.

Suppose to the contrary, and let $(S_1, S_2) \in T$. Define
$$
S_1' := \begin{cases}
S_1 &, |S_1| \geq 3, \\
L\setminus S_1 &, \mathrm{otherwise}.
\end{cases}
$$
Then $|S_1'| \geq 3$, and $(S_1', S_2) \in T$. (Verify!) Choose $B \subseteq L$ as follows. If $|S_1' \cap S_2| \geq 2$, let $B \subseteq S_1'\cap S_2$ be any subset of cardinality $2$. Otherwise, let $B \subseteq S_1' \setminus S_2$ be any subset of cardinality $2$. Then $\tau_B \in [I]_{S_1'} \cap [I]_{S_2}$, so that $\tau_B = I$, a contradiction!

Generalizations
The proof can be readily adjusted to account for an initial set of $n \geq 5$ cubes equipped with an arbitrary $n$-tuple, $(r_1, r_2, \dots, r_n)$, of distinct real numbers representing the cubes' "weights" (if the "weights" are not distinct, then it is trivially impossible to tell whether the labeling is correct by merely weighing the cubes).
A: Weigh the 2, 3, 5 against the 4, 6. If it doesn't balance, then you know there's an incorrect label. If it does balance, then either the labels are correct, or you're weighing 1, 2, 6 against 4, 5, or you're weighing 1, 3, 4 against 2, 6. Now weigh the 1, 2, 3 against the 6. This can't balance in the other two cases, only in the labels-correct case, and you're done.
