No.

Define $g : \mathbb{N} \times [0,1] \to \mathbb{R}$ by $f(n,x) := \operatorname{max}(0,1+((-n)\cdot \operatorname{inf}(\{|x+(-y)| : y\in (\operatorname{Cantor} \operatorname{set})\}))).$

$g$ is continuous, so it is in Baire class 0.

Define $f : [0,1] \to [0,1]$ by $f(x) := \displaystyle\lim_{n\to \infty} g(n,x)$.

$f$ is in Baire class 1, and is the characteristic function of the Cantor set.

$f$ is 0 almost everywhere, but not at all but countably many points.

$\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$.

Therefore there exists a Baire class 1 function $f$ such that $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$ and it is not the case that $f$ is zero at all but countably many points.

QED

If all discontinuities of $f$ are jump discontinuities, then the set of discontinuities of $f$ is countable, see http://en.wikipedia.org/wiki/Regulated_function. If $f$ additionally satisfies you integral condition, then $f$ is 0 almost everywhere, so in particular $f$ is 0 densely often, in which case $f$ is 0 everywhere it is continuous. Therefore, if $f : [0,1] \to \mathbb{R}$ satisfies $\int_0^1 f(x)p(x)dx=0$ for all polynomials $p$ and all discontinuities of $f$ are jump discontinuities, then $f$ is 0 at all but countably many points, and so in particular $f$ is the 0 function redefined at countably many points.

QED

Note that $f$ is not assumed to be in Baire class 1.