Suppose that $g_k\in C([1,2])$, $k\in \mathbb N$ are continuous functions such that $\|g_k\|_{C([1,2])} \leq \epsilon^k$ for some sufficiently small $\epsilon>0$. Is the following claim true:
If $f\in C([1,2])$ satisfies $$ \int_1^2 f(x)\, (x^n + g_n(x))\,dx =0 \quad \forall n\in \mathbb N,$$ then $f=0$ on $[1,2]$.
This is not an answer to the question, but a proof of something much weaker: there is a sequence $c_m\to0$ such that if $\lVert g_m\rVert\leq c_m$ for all $m$, then the result from the question is true. The growth of $c_m$ could be estimated if we had bounds for the constants $a_{m,i}$ and $k_m$ from below; I don't know if there are proofs of the Stone Weierstrass theorem including those bounds.
First note that for all $m$, the functions $x^m,x^{m+1},\dots$ generate a dense subspace of $C([1,2])$, as for any $h\in C([1,2])$ we can approximate $\frac{h(x)}{x^m}$ arbitrarily well with polynomials.
Now let $(f_m)_m$ be a sequence of continuous functions dense in $C([1,2])$. For each $f_m$, by the paragraph above we can obtain finitely many constants $a_{m,m+1},a_{m,m+2},\dots,a_{m,m+k_m}$ such that $$\left\| f_m-\sum_{i=m+1}^{m+k_m}a_{m,i}x^i\right\|_\infty<\frac{1}{m}.$$ So, if we consider a decreasing sequence $c_m$ such that $c_m\left|\sum_{i=m+1}^{m+k_m}a_{m,i}\right|\to0$, then $\lVert g_m\rVert<c_m\;\forall m$ implies that $$\left\|f_m-\sum_{i=m+1}^{m+k_m}a_{m,i}(x^i+g_i(x))\right\|_\infty\to0,$$ so the subspace of $C([a,b])$ generated by $x^i+g_i(x)$ is dense in $L^2$, proving that $f=0$.
Following Jochen Wengenroth suggestion, I sketch the proof of the result from a comment.
The claim follows from the following
Lemma. If $\mu$ is a measure supported on $[0,X]$, where $X>1$ such that the sequence $$\int_0^X x^n d\mu$$ is uniformly bounded, then $\mu$ is supported on $[0,1]$.
Proof: Suppose that $\left|\int_0^X x^n d\mu\right|\leq M$ for all $n$. Consider the Fourier transform $\hat{\mu}$. The bounds on integrals are equivalent to $|\hat{\mu}^{(n)}(0)|\leq M$. Thus $\hat{\mu}$ is an entire function of order 1 and type (not more than) 1. Also, it is bounded on a real line (by $C=\int d|\mu|$). By Phragmén–Lindelöf principle, $|\hat{\mu}(s+it)|\leq C e^{|t|}$. The lemma follows from the Paley-Wiener-Schwartz theorem (L. Hörmander "The analysis of linear partial differential operators", vol. I, Theorem 7.3.1).
It seems so. I follow the strategy suggested in the answer by Saúl RM.
Put $\epsilon=1/100000$.
We do not use the functions $g_n$ theirselves, only that $\int_0^1 f(x)x^n=O(\epsilon^n)$.
Lemma. There exists a polynomial $p_n(x)=\sum_{k=n}^{10 n} c_k x^k$ such that $\sum_{k=n}^{10n}|c_k|\epsilon^k=o(1)$ and $\max_{x\in [1,2]} |1-p_n(x)|=o(1)$.
First of all, I prove your claim using the polynomials from lemma. It suffices to prove that $\int_1^2 f(x)x^ddx=0$ for arbitrary non-negative integer $d$. We have $$ \int_1^2 f(x)x^d dx=\int_1^2 f(x)x^d(1-p_n(x))dx+\int_1^2 f(x)x^dp_n(x)dx\\ =o(1)+\sum_{k=n}^{10 n} c_k \int_1^2 f(x)x^{d+k}dx=o(1), $$ and the claim follows.
Proof of the lemma. Consider the Taylor approximation of degree $9n$ at point $3/2$ of the function $F(x):=x^{-n}$: $$ x^{-n}=\sum_{k=0}^{9n} \frac{n(n+1)\cdots (n+k-1)}{k!}(3/2)^{-n-k}(x-3/2)^k+R_{9n}(x), $$ where the remainder term $R_{9n}(x)=\frac{F^{(9n+1)(\theta)}}{(9n+1)!}(x-3/2)^{9n+1}$ for an intermediate point $\theta$ between $3/2$ and $x$ enjoys on $[1,2]$ the bound $$ |R_{9n}(x)|\leqslant (1/2)^{9n+1}\cdot \frac{n(n+1)\cdots (n+9n)}{(9n+1)!} =2^{-9n-1}{10n\choose n-1}\\<2^{-9n-1}\frac{(10n)^n}{n!}<(10 e\cdot 2^{-9})^n<3^{-n}, $$ thus $x^n R_{9n}(x)$ is uniformly small on $[1,2]$. Put $$p_n(x)=x^n\sum_{k=0}^{9n} \frac{n(n+1)\cdots (n+k-1)}{k!}(3/2)^{-n-k}(x-3/2)^k=\sum_{k=n}^{10n} c_n x^n.$$ We have $$ \sum_{k=n}^{10n} |c_n|\epsilon^n\leqslant \epsilon^n \sum_{k=0}^{9n} \frac{n(n+1)\cdots (n+k-1)}{k!}(3/2)^{-n-k}(\epsilon+3/2)^k \\<\epsilon^n (3/2)^{-n}\sum_{k=0}^{9n} {n+k-1\choose k}\cdot 2^k < \epsilon^n (3/2)^{-n}\sum_{k=0}^{9n} (1+2)^{n+k-1}=o(1) $$ as needed.
