The answer is No.
Let us assume that the claim you are asking about is true and let us try to arrive at a contradiction.
Cardinality of the set of all Borel measurable functions is $\mathfrak c=2^{\aleph^0}$; see here.
Let us well-order all positive Borel functions as $\{g_\alpha; \alpha<\mathfrak c\}$. Let us consider some well-ordering $\mathbb R=\{x_\alpha; \alpha<\mathfrak c\}$ of real numbers. Let us define a new function $f\colon\mathbb R\to(0,\infty)$ as $$f(x_\alpha)=\frac{g_\alpha(x_\alpha)}2.$$ Clearly, there is no $\alpha$ such that $g_\alpha\le f$. This contradicts the given claim.
