Let $H_1$ and $H_2$ be Hilbert spaces, and let $A : H_1 \to H_2$ be a Fredholm operator. Let $Q : H_2 \to H_1$ be a Fredholm parametrix for $A$, i.e. $AQ - 1$ and $QA - 1$ are compact operators. Assume that there is some positive integer $p$ so that $(AQ - 1)^p$ and $(QA - 1)^p$ are trace-class operators. Then, $$\mathrm{ind}(A) = \mathrm{tr}((QA - 1)^p) - \mathrm{tr}((AQ - 1)^p), $$ where the left hand side from above is the Fredholm index of $A$.
I am not exactly sure how to prove the above. I know about Fedosov's formula, which says that $$\mathrm{ind}(A) = \mathrm{tr}(AS - SA) = \mathrm{tr}(AS - 1) - \mathrm{tr}(SA - 1), $$ where $S$ is a Fredholm parametrix for $A$ for which $AS - 1$ and $SA - 1$ are finite-rank operators, but I am not sure how to use it for this problem (where $AQ - 1$ and $QA - 1$ are only compact).
I thought about using Lidskii's theorem, that the trace is equal to the (possibly infinite) sum of eigenvalues (with algebraic multiplicity taken into account). We have the relationship between the spectrums $$\sigma(AQ) \setminus \{0\} = \sigma(QA) \setminus \{0\}. $$ Moreover if $\lambda \neq 0$ is an eigenvalue of $AQ$, with eigenvector $v$, then $Qv$ is a $\lambda$-eigenvector for $QA$; the converse also holds, hence $AQ$ and $QA$ have the same eigenvalues, and I am inclined to say that the algebraic multiplicities of these eigenvalues are the same. Also, the above spectrum relation gives us that (I am a bit unsure of this) $$\sigma(AQ - 1) \setminus \{-1\} = \sigma(QA -1 ) \setminus \{-1\}. $$
The $(-1)$-eigenspace of $AQ - 1$ is just the kernel of $AQ$, and the $(-1)$-eigenspace of $QA - 1$ is the kernel of $QA$. So, the above and Lidskii's theorem would imply that $$\mathrm{tr}((QA - 1)^p) - \mathrm{tr}((AQ - 1)^p) = \sum_{i = 1}^{\dim(\ker(QA))} (-1)^p - \sum_{i = 1}^{\dim(\ker(AQ))} (-1)^p, $$ but the above series could be infinite and I am not sure how to proceed further.
I can answer the question in the affirmative with the hypotheses that $AQ-1$ and $QA-1$ are Schatten $p$-class operators, which is considerably stronger than just requiring $(AQ-1)^p$ and $(QA-1)^p$ to be trace-class. To get a taste for the difference of these, observe that an operator with a block form $$ T=\pmatrix{0 & I \cr 0 & 0}, $$ where $I$ is the identity on an infinite dimensional Hilbert space, satisfies $T^2=0$, but it is not in any Schatten $p$-class.
This said, let us prove the index formula $$ \text{ind}(A)= \text{tr}\big ((QA -1)^p - (AQ - 1)^p\big ) \tag{$\dagger$} $$ proposed by the OP with the stronger hypothesis referred to.
The first and most important step will be to show that the right-hand-side of ($\dagger$) does not depend on the parametrix $Q$ chosen.
In order to prove it, let us suppose that we have two parametrixes $Q$ and $Q'$, meaning, as observed above, that $AQ-1$, $QA-1$, $AQ'-1$, and $Q'A-1$, all belong to the Schatten $p$-class $L^p(H)$.
Recalling that $L^p(H)$ is an ideal in $B(H)$, the above says that the classes of both $Q$ and $Q'$ in $B(H)/L^p(H)$ serve as inverses for the class of $A$. Since inverses are unique, we deduce that $Q'\equiv Q$ (mod $L^p(H)$), so we may write $$ Q'=X+Q, $$ where $X\in L_p(H)$.
Our goal will therefore be to prove that $$ \text{tr}\big ((QA -1)^p-(AQ - 1)^p\big ) = \text{tr}\big ((Q'A -1)^p-(AQ' - 1)^p\big ).\tag{$*$} $$
If $a$ and $b$ are elements of a non-commutative algebra, and $p$ is a positive integer, we will need to express $(a+b)^p$ in terms of products of $a$ and $b$. Of course Newton's Binomial Theorem is not valid here, so we will employ a somewhat brute force replacement for it, namely we will consider the two-element set $F=\{a, b\}$ and for every $$ y\in F\times F\times \ldots \times F= F^p, $$ we may think of the product $y_1y_2\ldots y_p$ as one of the terms of our generalized Binomial Theorem which reads $$ (a+b)^p = \sum_{y\in F^p} y_1y_2\ldots y_p. $$ This said, consider the folowing two two-element sets $$ L=\{AX, AQ-1\} \text{ and } R=\{XA, QA-1\}, $$ so our Binomial Theorem says that $$ (AQ' - 1)^p = (AX+AQ - 1)^p = \sum_ {Y\in L^p} Y_1Y_2\ldots Y_p, $$ and $$ (Q'A - 1)^p = (XA+QA - 1)^p = \sum_ {Y\in R^p} Y_1Y_2\ldots Y_p. $$
For $Y\in L$, let us denote by $\bar Y$ the corresponding element of $R$, namely when $Y=AX$, we put $\bar Y=XA$, and when $Y=AQ-1$, then $\bar Y=QA-1$. It is then easy to see that $$ YA=A\bar Y, \ \forall Y\in L. $$
Given $Y\in L^p$, let us suppose that $Y$ has at least one coordinate equal to $AX$. We then claim that $$ \text{tr}(Y_1Y_2\ldots Y_p)=\text{tr}(\bar Y_1\bar Y_2\ldots \bar Y_p). $$ Indeed, assuming that the $i^{th}$ coordinate of $Y$ is $AX$, we have $$\def\quad{}\matrix{ \text{tr}(& Y_1& \ldots & Y_{i-1} & Y_i & Y_{i+1}& \ldots & Y_p&)= \cr \text{tr}(& Y_1& \ldots & Y_{i-1} & AX & Y_{i+1}& \ldots & Y_p&)= \cr \text{tr}(& Y_1& \ldots & A\bar Y_{i-1} & X & Y_{i+1}& \ldots & Y_p&)= \cr \text{tr}(& A\bar Y_1& \ldots & \bar Y_{i-1} & X & Y_{i+1}& \ldots & Y_p&)= \cr \text{tr}(& \bar Y_1& \ldots & \bar Y_{i-1} & X & Y_{i+1}& \ldots & Y_pA&)= \cr \text{tr}(& \bar Y_1& \ldots & \bar Y_{i-1} & X & Y_{i+1}& \ldots & A\bar Y_p&)= \cr \text{tr}(& \bar Y_1& \ldots & \bar Y_{i-1} & XA & \bar Y_{i+1}& \ldots & \bar Y_p&)= \cr \text{tr}(& \bar Y_1& \ldots & \bar Y_{i-1} & \bar Y_i & \bar Y_{i+1}& \ldots & \bar Y_p &).\hfill } $$
Denoting by $Z$ the element of $L^p$ whose coordinates are all equal to $AQ-1$, we then have that $$ \text{tr}\big ((AQ'-1)^p - (Q'A-1)^p\big ) = $$$$ = \text{tr}\big ((AX+AQ-1)^p - (XA+QA-1)^p\big ) = $$$$ = \text{tr}\Big ( (AQ-1)^p + \sum_ {Y\in L^p\setminus \{Z\}} Y_1Y_2\ldots Y_p - (QA-1)^p - \sum_ {Y\in R^p\setminus \{\bar Z\}} Y_1Y_2\ldots Y_p \Big ) = $$$$ = \text{tr}\Big ( (AQ-1)^p + \sum_ {Y\in L^p\setminus \{Z\}} \bar Y_1\bar Y_2\ldots \bar Y_p - (QA-1)^p - \sum_ {Y\in R^p\setminus \{\bar Z\}} Y_1Y_2\ldots Y_p \Big ) = $$$$ = \text{tr}\big ( (AQ-1)^p - (QA-1)^p\big ). $$
This proves ($*$).
Recal that a Fredholm operator $A$ admits a parametrix $P$ such that $AP$ is the orthogonal projection onto the range of $A$, and $1-PA$ is the orthogonal projection onto the kernel of $A$. If we then use $P$ in place of $Q$ in ($\dagger$), we easily reach the desired conclusion (perhaps up to a sign).