Yes, that statement is independent of the axioms of set theory.

Let $X$ be a compact *strong S-space*, that is a compact space such that $X^n$ is hereditarily separable but not hereditarily Lindelof for every $n \in \mathbb{N}$ (De La Vega and Kunen constructed a homogeneous space with these features in http://www.sciencedirect.com/science/article/pii/S0166864103002153)

If $C_p(X)$ contained an uncountable discrete set, this would yield an uncountable discrete set in $X^n$ for some $n$. But this is impossible, because every subspace of $X^n$ is separable. Now, $X$ is not metrizable, because separable metrizable spaces have a countable base, and hence are hereditarily Lindelof.

EDIT: It remains to prove that if $C_p(X)$ contains an uncountable discrete set then some finite power of $X$ also does. I believe this is due to Tkachuk.

Let $D=\{f_\alpha: \alpha <\omega_1\}$ be a discrete set in $C_p(X)$ of size $\omega_1$. Let $\mathcal{B}$ be a countable base for $\mathbb{R}$. Since $D$ is discrete, for every $\alpha < \omega_1$, we can find a positive integer $n_\alpha$, a sequence $(x^\alpha_1, \dots x^\alpha_{n_\alpha})$ of elements of $X$ and a sequence $(B^\alpha_1, \dots B^\alpha_{n_\alpha})$ of elements of $\mathcal{B}$ such that $[x^\alpha_1, \dots x^\alpha_{n_\alpha}, B^\alpha_1, \dots, B^\alpha_{n_\alpha}] \cap D=\{f_\alpha\}$, where $[x_1, \dots x_n, B_1, \dots B_n]$ denotes the following basic open subset of $C_p(X)$: $\{f \in C_p(X): (\forall i \leq n)(f(x_i) \in B_i)\}$.

By the pigeonhole principle, we can find an uncountable set $E \subset \omega_1$, a positive integer $n$ and a finite sequence $(B_1, \dots B_n)$ of elements of $\mathcal{B}$ such that $n_\alpha=n$ and $(B^\alpha_1, \dots B^\alpha_{n_\alpha})=(B_1, \dots B_n)$, for every $\alpha \in E$.

We claim that $S=\{(x^\alpha_1, \dots, x^\alpha_n): \alpha \in E \}$ is a discrete subset of $X^n$. Indeed, by definition of the $B_i$s and the $x^\beta_i$s, we have that $f_\beta$ is the only element of $D$ such that $f_\beta(x^\beta_i) \in B_i$ for every $i \leq n$. But this implies that the open set $\prod_{i=1}^n f_\alpha^{-1}(B_i)$ intersects $S$ in the single point $(x^\alpha_1, \dots x^\alpha_n)$, and hence $S$ is discrete in $X^n$.