Is the winding number of a map $f: S^k \to S^k$ that is discontinous even defined? What happens if one plugs in a discontinuous function $U$ inside the winding number formula? Is it an integer?
$$ W[U] = \frac{1}{24 \pi^2} \int d^3x \ \epsilon^{\mu \nu \beta} \ \text{Tr}(U \partial_{\mu} U^{-1} \ U \partial_{\nu} U^{-1} \ U \partial_{\beta} U^{-1})$$
Let me try to approach your question mathematically, pretending that you are a strong undergraduate math major who took an upper division real analysis sequence. This is the best I can do without teaching you such a sequence.
I suspect, what you have in mind is the notion of degree for maps $S\to S$, where $S=S^k$, the $k$-dimensional sphere. Analytically, degree is given by the formula $$ deg(f)= \frac{1}{\int_S \omega}\int_{S} f^*(\omega),$$ where $\omega$ is any volume form on the sphere $S$, i.e. a smooth form of degree $k$ which does not vanish at any point. (Judging by the expression in your integral including the presence of traces in the formula, you are using some particular volume form on the 3-dimensional sphere, presumably, coming from the identification of this sphere with the group $SU(2)$, where the volume for is the one defined using the Killing form of $SU(2)$. The normalization factor in front of the integral in your formula is likely to be the reciprocal of $\int_S \omega$ for your choice of $\omega$. The ratio of integrals in my formula, once properly defined, will be independent of the form $\omega$ and this will work in all dimensions.)
You need some analytical (in the sense of some condition coming from the real analysis) assumptions on $f$ for this integral to exist: Just the fact that you wrote a formula does not mean that it represents a valid mathematical expression. For instance, if $s=h(t)$ is a function of one real variable, the expression $$ \int_a^b h(t)dt $$ is not always meaningful, even if you allow for infinite values.
In local coordinates on the sphere, if $\omega=\phi(y)dy_1\wedge ...\wedge dy_k$, $y=f(x)$, then the pull-back form $f^{*}\omega$ is written as $$ I_f(x)dx_1\wedge ... \wedge dx_k= \phi(f(x)) J_f(x) dx_1\wedge ... \wedge dx_k, $$ where $J_f(x)$ is the Jacobian determinant of $f$ at the point $x$. For the integral to make sense you want $J_f(x)$ to exist except for a subset of measure zero and the function $I_f(x)=\phi(f(x)) J_f(x)$ to be (locally) integrable. Since $J_f$ is defined using 1st order partial derivatives, the most natural class of maps $f$ to consider is the Sobolev class $W^{1,k}_{loc}$.
Very roughly speaking (and it is literally wrong, but that's the best I can do in the context of the question) this means that $f=(f_1,...,f_k)$ and for each component function $f_i$, its 1st order partial derivatives $$ \frac{\partial f_i}{\partial x_j} $$ exist, except for a measure zero subset of the domain, and you can integrate these partials (over small $k$-dimensional balls) after multiplying by smooth functions.
A more honest way to describe functions $f_i$ in this Sobolev class is that (after altering $f_i$ on a subset of measure zero which is irrelevant for us) the function is (absolutely) continuous on almost every coordinate line, provided that partial derivatives above have integrable $k$-th power. The fact that one may want to require integralibity of the $k$-powers of partial derivatives can be heuristically justified by the fact that in the definition of the Jacobian determinant we are using $k$-fold products of partial derivatives.
Almost every for instance means that you can ignore all lines given by equations $x_j=q$, where $q$'s are rational numbers. I do not know if you are aware of this, but continuity on every line does not imply continuity of a function once you have more than one variable: Such a function can be discontinuous along some nonlinear curves, say, parabolas or spirals.
Absolutely continuous is a strong form of continuity which allows one to "do calculus." A function $s=h(t)$ of one real variable is absolutely continuous if it is continuous, has derivative almost everywhere in its domain (an interval) and satisfies the Fundamental Theorem of Calculus: $$ h(b)-h(a)= \int_a^b h'(t)dt $$ for all $a, b$ in the domain of $h$. (There are nonconstant continuous functions with derivative equal to zero almost everywhere, hence, failing absolute continuity.)
Thus, if your sphere is 1-dimensional, your map $f$ will be essentially continuous and you can as well work with continuous maps of circles. But in higher dimensions, one can allow for discontinuous maps with "controlled" discontinuity.
Regrading integrality: For maps $f$ in the Sobolev class $W^{1,k}$ the degree defined by the above integral formula will be an integer. One can prove this by using density of smooth maps in $W^{1,k}$ and then appealing to integrality of the ordinary degree.
Lastly: One of the best things about the notion of degree of a map is that it does not change under homotopy (a continuous deformation) of the map $f$. This property becomes meaningless if you allow discontinuous maps. So, what the definition of degree (given by the integral formula above) is good for once applied to discontinuous maps in the Sobolev class $W^{1,k}$, is very much unclear.
Remark. One example I do remember where the integral degree formula was used for "low-regularity" (but still continuous) maps $f$ in the Sobolev class $W^{1,k}_{loc}$ was some work of Bonk and Heinonen on quasiregular maps, proving a form of Picard's theorem. However, this is irrelevant for your question.
Here are references for people interested in math behind this staff:
Gol’dshtejn, V. M.; Kuz’minov, V. I.; Shvedov, I. A., De Rham isomorphism of the $L_p$-cohomology of noncompact Riemannian manifolds, Sib. Math. J. 29, No. 2, 190-197 (1988); translation from Sib. Mat. Zh. 29, No. 2(168), 34-44 (1988). ZBL0668.58051.
Slides of some random talk on this subject found by my search engine:
Nikolai Nowaczyk, The de Rham Isomorphism and the $L_p$-Cohomology of non-compact Riemannian Manifolds.