Can we not output the Kolmogorov complexity?
The question can be rephrased as whether or not $\lim \inf_{\vert x \vert \rightarrow \infty}{\vert T(x) - K(x) \vert} = 0$, and as Denis points out in the comments this is false for some encodings.  Here is a weaker statement and an attempted proof of it that doesn't depend on any details of the encoding, but I'll assume a binary language for simplicity:

Let $T:\{0,1\}^* \rightarrow \mathbb{N}$ be a computable function satisfying $0 \le T(x) \le \vert x \vert$ and $\lim \inf_{\vert x \vert \rightarrow \infty}{T(x)} = \infty$.  Then $\lim \inf_{\vert x \vert \rightarrow \infty}{\vert T(x) - K(x) \vert} \lt \infty$.  Informally, if there is a target around each string's Kolmogorov complexity that grows unboundedly wide, no computable function can avoid hitting it.

To see this, let $n$ be a random $b$-bit number, i.e. $0 \le n \lt 2^b$ and $K(n) \ge b$.  For all $b$ such a random $n$ exists.   Also note that there are an infinite number of values of $b$ for which $\vert \{T(x) = b\} \vert \ge 2^b$, this follows from the conditions placed on $T$.  Now let $x$ be the $n^{\text{th}}$ smallest string such that $T(x) = b$.  Clearly there is a constant $c_1$ such that $K(x) \gt b - c_1$, because $K(n) \ge b$ and $n$ can be computed from $x$.  And there is a constant $c_2$ such that $K(x) \lt b + c_2$, because $K(n)$ is also bounded from above by only a constant more than $b$, and $x$ can be computed from $n$.  Then $\vert K(x) - T(x) \vert \lt c_1 + c_2$, and we have an infinite number of choices for $b$ (those with a preimage of cardinality at least $2^b$), yielding an infinite number of values for $x$, so we are done.

An implication is that for some $c \in \mathbb{Z}$, $T(x) = K(x) + c$ infinitely often.  So one might say we can't not output something that's not the Kolmogorov complexity!